6.1 Trigonometry

Detailed Theory: Trigonometry

1. Basic Concepts and Angles

1.1 What is Trigonometry?

Trigonometry is the study of relationships between angles and sides of triangles.

Etymology: Greek words "trigonon" (triangle) + "metron" (measure)

1.2 Angle Measurement Systems

a) Degree Measure

  • Full circle = 360360^\circ

  • Right angle = 9090^\circ

  • Straight angle = 180180^\circ

b) Radian Measure

  • More natural for mathematics

  • Based on arc length

  • Full circle = 2π2\pi radians

  • 180=π180^\circ = \pi radians

c) Conversion Formulas

Degrees to Radians: θrad=π180×θdeg\text{Degrees to Radians: } \theta_{\text{rad}} = \frac{\pi}{180} \times \theta_{\text{deg}}
Radians to Degrees: θdeg=180π×θrad\text{Radians to Degrees: } \theta_{\text{deg}} = \frac{180}{\pi} \times \theta_{\text{rad}}

d) Common Conversions

30=π6 radians30^\circ = \frac{\pi}{6} \text{ radians}
45=π4 radians45^\circ = \frac{\pi}{4} \text{ radians}
60=π3 radians60^\circ = \frac{\pi}{3} \text{ radians}
90=π2 radians90^\circ = \frac{\pi}{2} \text{ radians}
180=π radians180^\circ = \pi \text{ radians}
360=2π radians360^\circ = 2\pi \text{ radians}

1.3 Types of Angles

  1. Acute angle: 0<θ<900^\circ < \theta < 90^\circ

  2. Right angle: θ=90\theta = 90^\circ

  3. Obtuse angle: 90<θ<18090^\circ < \theta < 180^\circ

  4. Straight angle: θ=180\theta = 180^\circ

  5. Reflex angle: 180<θ<360180^\circ < \theta < 360^\circ

  6. Complete angle: θ=360\theta = 360^\circ


2. Trigonometric Ratios

2.1 Right Triangle Trigonometry

Consider right triangle with:

  • Hypotenuse (opposite right angle)

  • Opposite side (opposite to angle θ\theta)

  • Adjacent side (next to angle θ\theta)

a) Basic Ratios

  1. Sine: sinθ=OppositeHypotenuse\sin\theta = \frac{\text{Opposite}}{\text{Hypotenuse}}

  2. Cosine: cosθ=AdjacentHypotenuse\cos\theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}

  3. Tangent: tanθ=OppositeAdjacent=sinθcosθ\tan\theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{\sin\theta}{\cos\theta}

b) Reciprocal Ratios

  1. Cosecant: cscθ=1sinθ=HypotenuseOpposite\csc\theta = \frac{1}{\sin\theta} = \frac{\text{Hypotenuse}}{\text{Opposite}}

  2. Secant: secθ=1cosθ=HypotenuseAdjacent\sec\theta = \frac{1}{\cos\theta} = \frac{\text{Hypotenuse}}{\text{Adjacent}}

  3. Cotangent: cotθ=1tanθ=AdjacentOpposite=cosθsinθ\cot\theta = \frac{1}{\tan\theta} = \frac{\text{Adjacent}}{\text{Opposite}} = \frac{\cos\theta}{\sin\theta}

2.2 Mnemonic for Remembering

SOH-CAH-TOA:

  • Sine = Opposite/Hypotenuse

  • Cosine = Adjacent/Hypotenuse

  • Tangent = Opposite/Adjacent

2.3 Values for Standard Angles

Angle = 3030^\circ or π6\frac{\pi}{6}

sin30=12\sin 30^\circ = \frac{1}{2}
cos30=32\cos 30^\circ = \frac{\sqrt{3}}{2}
tan30=13\tan 30^\circ = \frac{1}{\sqrt{3}}

Angle = 4545^\circ or π4\frac{\pi}{4}

sin45=12\sin 45^\circ = \frac{1}{\sqrt{2}}
cos45=12\cos 45^\circ = \frac{1}{\sqrt{2}}
tan45=1\tan 45^\circ = 1

Angle = 6060^\circ or π3\frac{\pi}{3}

sin60=32\sin 60^\circ = \frac{\sqrt{3}}{2}
cos60=12\cos 60^\circ = \frac{1}{2}
tan60=3\tan 60^\circ = \sqrt{3}

2.4 Trigonometric Table (0° to 90°)

Angle

sin\sin

cos\cos

tan\tan

00^\circ

0

1

0

3030^\circ

12\frac{1}{2}

32\frac{\sqrt{3}}{2}

13\frac{1}{\sqrt{3}}

4545^\circ

12\frac{1}{\sqrt{2}}

12\frac{1}{\sqrt{2}}

1

6060^\circ

32\frac{\sqrt{3}}{2}

12\frac{1}{2}

3\sqrt{3}

9090^\circ

1

0

\infty


3. Trigonometric Identities

3.1 Pythagorean Identities

  1. sin2θ+cos2θ=1\sin^2\theta + \cos^2\theta = 1

  2. 1+tan2θ=sec2θ1 + \tan^2\theta = \sec^2\theta

  3. 1+cot2θ=csc2θ1 + \cot^2\theta = \csc^2\theta

3.2 Reciprocal Identities

  1. cscθ=1sinθ\csc\theta = \frac{1}{\sin\theta}

  2. secθ=1cosθ\sec\theta = \frac{1}{\cos\theta}

  3. cotθ=1tanθ\cot\theta = \frac{1}{\tan\theta}

3.3 Quotient Identities

  1. tanθ=sinθcosθ\tan\theta = \frac{\sin\theta}{\cos\theta}

  2. cotθ=cosθsinθ\cot\theta = \frac{\cos\theta}{\sin\theta}

3.4 Co-function Identities

  1. sin(90θ)=cosθ\sin(90^\circ - \theta) = \cos\theta

  2. cos(90θ)=sinθ\cos(90^\circ - \theta) = \sin\theta

  3. tan(90θ)=cotθ\tan(90^\circ - \theta) = \cot\theta

  4. cot(90θ)=tanθ\cot(90^\circ - \theta) = \tan\theta

  5. sec(90θ)=cscθ\sec(90^\circ - \theta) = \csc\theta

  6. csc(90θ)=secθ\csc(90^\circ - \theta) = \sec\theta

3.5 Even-Odd Identities

  1. sin(θ)=sinθ\sin(-\theta) = -\sin\theta (odd function)

  2. cos(θ)=cosθ\cos(-\theta) = \cos\theta (even function)

  3. tan(θ)=tanθ\tan(-\theta) = -\tan\theta (odd function)

3.6 Periodicity Identities

  1. sin(θ+2π)=sinθ\sin(\theta + 2\pi) = \sin\theta

  2. cos(θ+2π)=cosθ\cos(\theta + 2\pi) = \cos\theta

  3. tan(θ+π)=tanθ\tan(\theta + \pi) = \tan\theta


4. Trigonometric Functions of Any Angle

4.1 Unit Circle Approach

Unit circle: Circle with radius 1 centered at origin

For angle θ\theta measured counterclockwise from positive x-axis:

  • Point on circle: (cosθ,sinθ)(\cos\theta, \sin\theta)

  • x-coordinate = cosθ\cos\theta

  • y-coordinate = sinθ\sin\theta

4.2 Signs in Different Quadrants

Quadrant

sin\sin

cos\cos

tan\tan

I (0°-90°)

+

+

+

II (90°-180°)

+

-

-

III (180°-270°)

-

-

+

IV (270°-360°)

-

+

-

Mnemonic: "All Students Take Calculus"

  • All: All positive in Quadrant I

  • Students: Sine positive in Quadrant II

  • Take: Tangent positive in Quadrant III

  • Calculus: Cosine positive in Quadrant IV

4.3 Reference Angles

Reference angle = Acute angle between terminal side and x-axis

Finding reference angle α\alpha:

  • Quadrant I: α=θ\alpha = \theta

  • Quadrant II: α=180θ\alpha = 180^\circ - \theta

  • Quadrant III: α=θ180\alpha = \theta - 180^\circ

  • Quadrant IV: α=360θ\alpha = 360^\circ - \theta

Using reference angle:

sinθ=±sinα\sin\theta = \pm\sin\alpha
cosθ=±cosα\cos\theta = \pm\cos\alpha
tanθ=±tanα\tan\theta = \pm\tan\alpha

Sign depends on quadrant.

4.4 Example: Find sin210\sin 210^\circ

Step 1: 210210^\circ is in Quadrant III (180°-270°)

Step 2: Reference angle = 210180=30210^\circ - 180^\circ = 30^\circ

Step 3: In Quadrant III, sine is negative

Step 4: sin210=sin30=12\sin 210^\circ = -\sin 30^\circ = -\frac{1}{2}


5. Sum and Difference Formulas

5.1 Sine Formulas

sin(A+B)=sinAcosB+cosAsinB\sin(A + B) = \sin A \cos B + \cos A \sin B
sin(AB)=sinAcosBcosAsinB\sin(A - B) = \sin A \cos B - \cos A \sin B

5.2 Cosine Formulas

cos(A+B)=cosAcosBsinAsinB\cos(A + B) = \cos A \cos B - \sin A \sin B
cos(AB)=cosAcosB+sinAsinB\cos(A - B) = \cos A \cos B + \sin A \sin B

5.3 Tangent Formulas

tan(A+B)=tanA+tanB1tanAtanB\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}
tan(AB)=tanAtanB1+tanAtanB\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}

5.4 Double Angle Formulas

sin2A=2sinAcosA\sin 2A = 2 \sin A \cos A
cos2A=cos2Asin2A=2cos2A1=12sin2A\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A
tan2A=2tanA1tan2A\tan 2A = \frac{2\tan A}{1 - \tan^2 A}

5.5 Half Angle Formulas

sinA2=±1cosA2\sin\frac{A}{2} = \pm\sqrt{\frac{1 - \cos A}{2}}
cosA2=±1+cosA2\cos\frac{A}{2} = \pm\sqrt{\frac{1 + \cos A}{2}}
tanA2=±1cosA1+cosA=1cosAsinA=sinA1+cosA\tan\frac{A}{2} = \pm\sqrt{\frac{1 - \cos A}{1 + \cos A}} = \frac{1 - \cos A}{\sin A} = \frac{\sin A}{1 + \cos A}

5.6 Triple Angle Formulas

sin3A=3sinA4sin3A\sin 3A = 3\sin A - 4\sin^3 A
cos3A=4cos3A3cosA\cos 3A = 4\cos^3 A - 3\cos A
tan3A=3tanAtan3A13tan2A\tan 3A = \frac{3\tan A - \tan^3 A}{1 - 3\tan^2 A}

6. Product-to-Sum and Sum-to-Product Formulas

6.1 Product-to-Sum Formulas

sinAcosB=12[sin(A+B)+sin(AB)]\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]
cosAsinB=12[sin(A+B)sin(AB)]\cos A \sin B = \frac{1}{2}[\sin(A+B) - \sin(A-B)]
cosAcosB=12[cos(A+B)+cos(AB)]\cos A \cos B = \frac{1}{2}[\cos(A+B) + \cos(A-B)]
sinAsinB=12[cos(A+B)cos(AB)]\sin A \sin B = -\frac{1}{2}[\cos(A+B) - \cos(A-B)]

6.2 Sum-to-Product Formulas

sinA+sinB=2sin(A+B2)cos(AB2)\sin A + \sin B = 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)
sinAsinB=2cos(A+B2)sin(AB2)\sin A - \sin B = 2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)
cosA+cosB=2cos(A+B2)cos(AB2)\cos A + \cos B = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)
cosAcosB=2sin(A+B2)sin(AB2)\cos A - \cos B = -2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)

7. Trigonometric Equations

7.1 Basic Solution Method

To solve sinθ=k\sin\theta = k or cosθ=k\cos\theta = k or tanθ=k\tan\theta = k:

  1. Find principal solution (using inverse functions)

  2. Use periodicity to find general solution

7.2 General Solutions

a) For sinθ=sinα\sin\theta = \sin\alpha

θ=nπ+(1)nα,nZ\theta = n\pi + (-1)^n\alpha, \quad n \in \mathbb{Z}

b) For cosθ=cosα\cos\theta = \cos\alpha

θ=2nπ±α,nZ\theta = 2n\pi \pm \alpha, \quad n \in \mathbb{Z}

c) For tanθ=tanα\tan\theta = \tan\alpha

θ=nπ+α,nZ\theta = n\pi + \alpha, \quad n \in \mathbb{Z}

7.3 Examples

Example 1: Solve sinθ=12\sin\theta = \frac{1}{2}

Solution:

Reference angle: sin30=12\sin 30^\circ = \frac{1}{2}

General solution:

θ=nπ+(1)nπ6,nZ\theta = n\pi + (-1)^n \frac{\pi}{6}, \quad n \in \mathbb{Z}

Specific solutions in [0,2π)[0, 2\pi): π6\frac{\pi}{6}, 5π6\frac{5\pi}{6}

Example 2: Solve cosθ=12\cos\theta = -\frac{1}{2}

Solution:

Reference angle: cos60=12\cos 60^\circ = \frac{1}{2}

Since cosine is negative in Quadrants II and III:

In Quadrant II: θ=18060=120\theta = 180^\circ - 60^\circ = 120^\circ or 2π3\frac{2\pi}{3}

In Quadrant III: θ=180+60=240\theta = 180^\circ + 60^\circ = 240^\circ or 4π3\frac{4\pi}{3}

General solution: θ=2nπ±2π3\theta = 2n\pi \pm \frac{2\pi}{3}


8. Graphs of Trigonometric Functions

8.1 Sine Function: y=sinxy = \sin x

  • Domain: All real numbers

  • Range: [1,1][-1, 1]

  • Period: 2π2\pi

  • Amplitude: 1

  • Zeros: x=nπx = n\pi, nZn \in \mathbb{Z}

  • Maximum: 11 at x=π2+2nπx = \frac{\pi}{2} + 2n\pi

  • Minimum: 1-1 at x=3π2+2nπx = \frac{3\pi}{2} + 2n\pi

8.2 Cosine Function: y=cosxy = \cos x

  • Domain: All real numbers

  • Range: [1,1][-1, 1]

  • Period: 2π2\pi

  • Amplitude: 1

  • Zeros: x=π2+nπx = \frac{\pi}{2} + n\pi

  • Maximum: 11 at x=2nπx = 2n\pi

  • Minimum: 1-1 at x=π+2nπx = \pi + 2n\pi

8.3 Tangent Function: y=tanxy = \tan x

  • Domain: xπ2+nπx \neq \frac{\pi}{2} + n\pi

  • Range: All real numbers

  • Period: π\pi

  • Vertical asymptotes: x=π2+nπx = \frac{\pi}{2} + n\pi

  • Zeros: x=nπx = n\pi

8.4 General Form: y=Asin(Bx+C)+Dy = A\sin(Bx + C) + D

  • A: Amplitude (vertical stretch)

  • B: Affects period (Period = 2πB\frac{2\pi}{|B|})

  • C: Phase shift (horizontal shift = CB-\frac{C}{B})

  • D: Vertical shift

8.5 Example: Graph y=2sin(3xπ)+1y = 2\sin(3x - \pi) + 1

Amplitude: 22

Period: 2π3\frac{2\pi}{3}

Phase shift: π3\frac{\pi}{3} to the right

Vertical shift: 11 up

Range: [1,3][-1, 3]


9. Inverse Trigonometric Functions

9.1 Definitions and Ranges

a) Inverse Sine: y=sin1xy = \sin^{-1}x or y=arcsinxy = \arcsin x

  • Domain: [1,1][-1, 1]

  • Range: [π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

b) Inverse Cosine: y=cos1xy = \cos^{-1}x or y=arccosxy = \arccos x

  • Domain: [1,1][-1, 1]

  • Range: [0,π][0, \pi]

c) Inverse Tangent: y=tan1xy = \tan^{-1}x or y=arctanxy = \arctan x

  • Domain: All real numbers

  • Range: (π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

9.2 Important Properties

  1. sin1(sinx)=x\sin^{-1}(\sin x) = x only if x[π2,π2]x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

  2. sin(sin1x)=x\sin(\sin^{-1} x) = x for x[1,1]x \in [-1, 1]

  3. cos1(x)=πcos1x\cos^{-1}(-x) = \pi - \cos^{-1}x

  4. sin1x+cos1x=π2\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}

  5. tan1x+cot1x=π2\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}

  6. sec1x+csc1x=π2\sec^{-1}x + \csc^{-1}x = \frac{\pi}{2}

9.3 Example: Find sin1(12)\sin^{-1}\left(\frac{1}{2}\right)

We need angle in [π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] with sine = 12\frac{1}{2}

sin1(12)=π6\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}

10. Applications of Trigonometry

10.1 Solving Triangles

a) Law of Sines

For any triangle with sides a,b,ca, b, c opposite angles A,B,CA, B, C:

asinA=bsinB=csinC=2R\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R

where RR is circumradius.

b) Law of Cosines

a2=b2+c22bccosAa^2 = b^2 + c^2 - 2bc\cos A
b2=a2+c22accosBb^2 = a^2 + c^2 - 2ac\cos B
c2=a2+b22abcosCc^2 = a^2 + b^2 - 2ab\cos C

c) Law of Tangents

aba+b=tan(AB2)tan(A+B2)\frac{a-b}{a+b} = \frac{\tan\left(\frac{A-B}{2}\right)}{\tan\left(\frac{A+B}{2}\right)}

10.2 Area Formulas

  1. Using base and height: A=12bhA = \frac{1}{2}bh

  2. Using two sides and included angle: A=12absinCA = \frac{1}{2}ab\sin C

  3. Heron's formula: A=s(sa)(sb)(sc)A = \sqrt{s(s-a)(s-b)(s-c)} where s=a+b+c2s = \frac{a+b+c}{2}

10.3 Example: Solve triangle with a=5a=5, b=6b=6, C=60C=60^\circ

Step 1: Find side cc using Law of Cosines:

c2=a2+b22abcosC=25+362(5)(6)cos60c^2 = a^2 + b^2 - 2ab\cos C = 25 + 36 - 2(5)(6)\cos 60^\circ
c2=6160×12=6130=31c^2 = 61 - 60 \times \frac{1}{2} = 61 - 30 = 31
c=315.57c = \sqrt{31} \approx 5.57

Step 2: Find angle AA using Law of Sines:

sinAa=sinCc\frac{\sin A}{a} = \frac{\sin C}{c}
sinA=asinCc=5×sin6031=5×3231=53231\sin A = \frac{a\sin C}{c} = \frac{5 \times \sin 60^\circ}{\sqrt{31}} = \frac{5 \times \frac{\sqrt{3}}{2}}{\sqrt{31}} = \frac{5\sqrt{3}}{2\sqrt{31}}
A=sin1(53231)A = \sin^{-1}\left(\frac{5\sqrt{3}}{2\sqrt{31}}\right)

Step 3: Find angle BB:

B=180ACB = 180^\circ - A - C

11. Trigonometric Series and Complex Numbers

11.1 Euler's Formula

eiθ=cosθ+isinθe^{i\theta} = \cos\theta + i\sin\theta

11.2 De Moivre's Theorem

For any integer nn:

(cosθ+isinθ)n=cos(nθ)+isin(nθ)(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)

11.3 Trigonometric Form of Complex Number

For complex number z=x+iyz = x + iy:

z=r(cosθ+isinθ)=reiθz = r(\cos\theta + i\sin\theta) = re^{i\theta}

where r=x2+y2r = \sqrt{x^2 + y^2} and θ=tan1(yx)\theta = \tan^{-1}\left(\frac{y}{x}\right)

11.4 Applications

  1. Finding nn-th roots: nn-th roots of z=r(cosθ+isinθ)z = r(\cos\theta + i\sin\theta) are:

    zk=r1/n[cos(θ+2kπn)+isin(θ+2kπn)]z_k = r^{1/n}\left[\cos\left(\frac{\theta + 2k\pi}{n}\right) + i\sin\left(\frac{\theta + 2k\pi}{n}\right)\right]

    for k=0,1,,n1k = 0, 1, \ldots, n-1

  2. Expressing cosnθ\cos^n\theta and sinnθ\sin^n\theta in terms of multiple angles


12. Trigonometric Inequalities

12.1 Solving Basic Inequalities

Example: Solve sinx>12\sin x > \frac{1}{2} for x[0,2π)x \in [0, 2\pi)

Solution:

From unit circle: sinx=12\sin x = \frac{1}{2} at x=π6x = \frac{\pi}{6} and x=5π6x = \frac{5\pi}{6}

Since sine is positive in Quadrants I and II:

Solution: π6<x<5π6\frac{\pi}{6} < x < \frac{5\pi}{6}

12.2 General Method

  1. Solve corresponding equation

  2. Identify intervals where inequality holds

  3. Consider periodic nature


13. Important Formulas Summary

13.1 Pythagorean Identities

sin2θ+cos2θ=1\sin^2\theta + \cos^2\theta = 1
1+tan2θ=sec2θ1 + \tan^2\theta = \sec^2\theta
1+cot2θ=csc2θ1 + \cot^2\theta = \csc^2\theta

13.2 Sum and Difference Formulas

sin(A±B)=sinAcosB±cosAsinB\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B
cos(A±B)=cosAcosBsinAsinB\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B

13.3 Double Angle Formulas

sin2A=2sinAcosA\sin 2A = 2\sin A \cos A
cos2A=cos2Asin2A\cos 2A = \cos^2 A - \sin^2 A

13.4 Half Angle Formulas

sinA2=±1cosA2\sin\frac{A}{2} = \pm\sqrt{\frac{1 - \cos A}{2}}
cosA2=±1+cosA2\cos\frac{A}{2} = \pm\sqrt{\frac{1 + \cos A}{2}}

13.5 Law of Sines and Cosines

asinA=bsinB=csinC\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}
a2=b2+c22bccosAa^2 = b^2 + c^2 - 2bc\cos A

14. Solved Examples

Example 1: Simplify sin2θcos2θsinθcosθ\frac{\sin^2\theta - \cos^2\theta}{\sin\theta \cos\theta}

Solution:

Using identities: sin2θcos2θ=cos2θ\sin^2\theta - \cos^2\theta = -\cos 2\theta

Also: sinθcosθ=12sin2θ\sin\theta\cos\theta = \frac{1}{2}\sin 2\theta

So:

sin2θcos2θsinθcosθ=cos2θ12sin2θ=2cot2θ\frac{\sin^2\theta - \cos^2\theta}{\sin\theta \cos\theta} = \frac{-\cos 2\theta}{\frac{1}{2}\sin 2\theta} = -2\cot 2\theta

Example 2: Prove 1+sinθcosθ=cosθ1sinθ\frac{1 + \sin\theta}{\cos\theta} = \frac{\cos\theta}{1 - \sin\theta}

Solution:

Cross multiply: (1+sinθ)(1sinθ)=cos2θ(1 + \sin\theta)(1 - \sin\theta) = \cos^2\theta

Left side: 1sin2θ=cos2θ1 - \sin^2\theta = \cos^2\theta (using sin2θ+cos2θ=1\sin^2\theta + \cos^2\theta = 1)

Right side: cos2θ\cos^2\theta

Both sides equal, identity proved.

Example 3: Solve 2cos2x3cosx+1=02\cos^2x - 3\cos x + 1 = 0 for 0x<2π0 \leq x < 2\pi

Solution:

Let t=cosxt = \cos x, then 2t23t+1=02t^2 - 3t + 1 = 0

Factor: (2t1)(t1)=0(2t - 1)(t - 1) = 0

So t=12t = \frac{1}{2} or t=1t = 1

Case 1: cosx=12\cos x = \frac{1}{2} Solutions: x=π3x = \frac{\pi}{3}, x=5π3x = \frac{5\pi}{3}

Case 2: cosx=1\cos x = 1 Solution: x=0x = 0

Total solutions: x=0x = 0, π3\frac{\pi}{3}, 5π3\frac{5\pi}{3}


15. Exam Tips and Common Mistakes

15.1 Common Mistakes

  1. Confusing degrees and radians: Always check which is being used

  2. Forgetting domain restrictions for inverse functions

  3. Sign errors when using reference angles

  4. Not considering all solutions in trigonometric equations

  5. Misapplying identities (especially with signs)

15.2 Problem-Solving Strategy

  1. Identify knowns and unknowns

  2. Choose appropriate identities/formulas

  3. Simplify step by step

  4. Check domain/range restrictions

  5. Verify solution when possible

15.3 Important Values to Memorize

  • sin0=0\sin 0^\circ = 0, cos0=1\cos 0^\circ = 1, tan0=0\tan 0^\circ = 0

  • sin30=12\sin 30^\circ = \frac{1}{2}, cos30=32\cos 30^\circ = \frac{\sqrt{3}}{2}, tan30=13\tan 30^\circ = \frac{1}{\sqrt{3}}

  • sin45=12\sin 45^\circ = \frac{1}{\sqrt{2}}, cos45=12\cos 45^\circ = \frac{1}{\sqrt{2}}, tan45=1\tan 45^\circ = 1

  • sin60=32\sin 60^\circ = \frac{\sqrt{3}}{2}, cos60=12\cos 60^\circ = \frac{1}{2}, tan60=3\tan 60^\circ = \sqrt{3}

  • sin90=1\sin 90^\circ = 1, cos90=0\cos 90^\circ = 0, tan90=undefined\tan 90^\circ = \text{undefined}

This comprehensive theory covers all aspects of trigonometry with detailed explanations and examples, providing complete preparation for the entrance examination.

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