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**Basic Trigonometric Identities**

- sin² θ + cos² θ = 1 ; -1 ≤ sin θ ≤ 1 ; -1 ≤ cos θ ≤ 1 , θ ∈ R
- sec² θ - tan² θ = 1 ; |sec θ| ≥ 1, θ ∈ R
- cosec² θ - cot² θ = 1 ; |cosec θ| ≥ 1 , θ ∈ R

- sin nπ = 0 ; cos nπ = (-1)
^{n}; tan nπ = 0 where n ∈ I

- If θ is any angle , then - θ, 90 ±θ , 180 ± θ, 270 ± θ , 360 ± θ etc. are called ALLIED ANGLES.

**Example ****1. Express 1·2 radians in degree measure.**

1·2 radians =

**Example ****2. Calculate sin α if cos α = - 9/11 and α ∈ (π, 3π/2).**

For any angle a belonging to the indicated interval sin a is negative, and therefore,

**Example 3****. Calculate tan α if cos α = √5/5 and α ∈ (π, 3π/2)**

For any angle a belonging to the indicated interval tanα is positive and cos a is negative, and therefore

- sin (A ± B) = sinA cosB ± cosA sinB
- cos (A ± B) = cosA cosB m sinA sinB
- sin²A - sin²B = cos²B - cos²A = sin (A+B) . sin (A- B)
- cos²A - sin²B = cos²B - sin²A = cos (A+B) . cos (A - B)

**➢ Factorisation of the Sum Or Difference of Two sines Or cosines:**

- sinC + sinD = 2 sin (C + D/2) cos (C - D/2)
- sinC - sinD = 2 cos (C + D/2) sin (C - D/2)
- cosC + cosD = 2 cos (C + D/2) cos (C - D/2)
- cosC - cosD = -
^{ }2 sin (C + D/2) sin (C - D/2)

**➢ Transformation of Products Into Sum Or Difference Of sines & cosines:**

- 2sinAcosB = sin(A+B) + sin(A-B)
- 2cosAsinB = sin(A+B) - sin(A-B)
- 2cosAcosB = cos(A+B) + cos(A-B)
- 2sinAsinB = cos(A-B) - cos(A+B)

**Example ****4. Suppose x and y are real numbers such that tan x + tan y = 42 and cot x + cot y = 49. Find the value of tan(x + y).**

tan x + tan y = 42 and cot x + cot y = 49

now, cot x + cot y = 49

**Example ****5.** If x sinθ = y sin(θ + 2π/3) = z sin (θ + 4π/3) then:**(a) x + y + z = 0 ****(b) xy + yz + zx = 0 ****(c) xyz + x + y + z = 1 ****(d) none**

Correct Answer is Option (b)= xy + yz + zx = 0

**Example ****6. ****Find θ satisfying the equation, tan 15° . tan 25° . tan 35° = tan 0, where θ ∈ (0, 15°). **

LHS = tan 15° . tan (30° - 5°) . tan (30° + 5°)

let t = tan 30° and m = tan 5°

**Example ****7.** **If tan A & tan B are the roots of the quadratic equation, a ^{ }x^{2} + b^{ }x + c = 0 then evaluate **

Now E = cos^{2}(A + B) [a tan^{2}(A + B) + b tan (A + B) + c]

**Example ****8. _{ }Show that cos^{2}A + cos^{2}(A + B) + 2 cosA cos(180° + B) · cos(360° + A + B) is independent of A. Hence find its value when B = 810°.**

cos

^{2}A + cos^{2}(A + B) - [2 cosA · cosB · cos (A + B)]

= cos^{2}A + cos^{2}(A + B) - [ {cos(A + B) + cos(A - B) } cos (A + B) ]

= cos^{2}A + cos^{2}(A + B) - cos^{2}(A + B) - (cos^{2}A - sin^{2}B)

= sin^{2}B which is independent of A

now, sin^{2}(810°) = sin^{2}(720° + 90°) = sin^{2}90° = 1

**Example ****9.** **Simplify: cos x · sin(y – z) + cos y · sin(z – x) + cos z · sin (x – y) where x, y, z ∈ R**

(1/2)[sin(y – z + x) + sin(y – z – x) + sin(z – x + y) + sin(z – x – y) + sin(x – y + z) + sin(x – y – z)] = 0

- sin 2A = 2 sinA cosA ; sin θ = 2 Sinθ/2 Cosθ/2
- cos 2A = cos²A - sin²A = 2cos²A - 1 = 1 - 2 sin²A ;

cos θ= cos² θ/2 - sin² θ/2 = 2cos² θ/2 - 1 = 1 - 2sin² θ/2.

2 cos²A = 1 + cos 2A , 2sin²A = 1 - cos 2A

2 cos² θ/2 = 1 + cos θ, 2 sin² θ/2 = 1 - cos θ. - cos 3A = 4 cos
^{3}A - 3 cosA - sin 3A = 3 sinA - 4 sin
^{3}A

If A+B+C = π

- tanA + tanB + tanC = tanA tanB tanC
- sin2A + sin2B + sin2C = 4 sinA sinB sinC
- sinA + sinB + sinC = 4 cos A/2 Cos B/2 Cos C/2
- cos 2A + cos 2B + cos 2C = –1 – 4 cos A cos B cos C
- cos A + cos B + cos C = 1 + 4 sin A/2 sin B/2 sin C/2

**Example ****10.** **If A + B + C = π, prove that,**

**Example ****11. ****If A + B + C = θ and cotθ = cot A + cot B + cot C, show that , sin(A-θ). sin (B-θ).sin (C-θ)= sin ^{3} θ.**

Given cot

θ= cot A + cot B + cot C or cotθ- cot A = cot B + cot C....(1)

....(2)

....(3)

Multiplying (1) , (2) and (3) we get the result

**Example ****12.** **Find whether a triangle ABC can exists with the tangents of its interior angle satisfying, tan A = x, tan B = x + 1 and tan C = 1 - x for some real value of x. Justify your assertion with adequate reasoning.**

In a triangle ∑ tan A = π tan A (to be proved)

x + x + 1 + 1 – x = x(1 + x)(1 – x)

2 + x = x – x^{3}; x^{3}= –2, x = -2^{1/3}

Hence tanA = x < 0 and tanB = x + 1 = 1 – 2^{1/3}< 0

Hence A and B both are obtuse. Which is not possible in a triangle. Hence no such triangle can exist.

- Min. value of a
^{2}tan^{2}θ + b^{2}cot^{2}θ = 2ab - Max and Min. value of acosθ + bsinθ are
- If f(θ) = acos(α + q) + bcos(β + q) where a, b, α and β are known quantities then
- (constant) then the maximum values of the expression cosα cosβ, cosα+ cosβ, sinα + sinβ and sina sinb occurs when α = β = σ/2
- If A, B, C are the anlges of a triangle then maximum value of sinA + sinB + sinC and sinA sinB sinC occurs when A = B = C = 60º
- In case a quadratic in sinq or cosq is given then the maximum or minimum values can be interpreted by making a perfect square

**Example ****13. _{ }Find the greatest value of c such that system of equations x^{2} + y^{2} = 25; x + y = c has a real solution.**

Put x = 5 cosθ and y = 5 sinθ

5(cosθ+ sinθ) = c; but (cosθ + sinθ)max = √2 and (cosθ + sinθ)min = – √2

hence, c

_{max}= 5√2

**Example ****14.** **Find the minimum and maximum value of f (x, y) = 7x^{2} + 4xy + 3y^{2} subjected to x^{2} + y^{2} = 1.**

Let x = cosθ and y = sinθ

y = f (θ) = 7 cos2θ + 4 sin θcosθ + 3 sin2θ = 3 + 2 sin 2q + 2(1 + cos 2θ)

= 5 + 2(sin 2θ + cos 2θ) but -√2 ≤ (sin 2θ + cos 2θ) ≤ √2

y_{max}= 5 + 2√2 and y_{min}= 5 – 2√2

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