Compound Angle Practice Questions for Grade 11 :
Here we are going to see some practice questions on based on compound angles.
(1) If sin x = 15/17 and cos y = 12/13 , 0 < x < π/2 , 0 < y < π/2, find the value of
(i) sin(x + y) (ii) cos(x − y) (iii) tan(x + y).
(2) If sin A = 3/5 and cos B = 9/41 , 0 < A < π/2, 0 < B < π/2, find the value of (i) sin(A + B) (ii) cos(A − B).
(3) Find cos(x − y), given that cos x = −4/5 with π < x < 3π/2 and sin y = −24/25 with π < y < 3π/2. Solution
(4) Find sin(x − y), given that sin x = 8/17 with 0 < x < π/2 and cos y = −24/25 with π < y < 3π/2 Solution
(5) Find the value of (i) cos 105° (ii) sin 105° (iii) tan 7π/12. Solution
(6) Prove that (i) cos(30° + x) = (√3 cos x − sin x)/2
(ii) cos(π + θ) = −cos θ
(iii) sin(π + θ) = −sin θ. Solution
(7) Find a quadratic equation whose roots are sin 15° and cos 15°. Solution
(8) Expand cos(A + B + C). Hence prove that
cosAcosB cosC = sinAsinB cosC + sinB sinC cosA + sinC sinAcos B, if A + B + C = π/2 Solution
(9) Prove that
(i) sin(45° + θ) − sin(45° − θ) = √2 sinθ. Solution
(ii) sin(30° + θ) + cos(60° + θ) = cos θ. Solution
(10) If a cos(x + y) = b cos(x − y), show that (a + b) tanx = (a − b) cot y. Solution
(11) Prove that sin 105° + cos 105° = cos 45° Solution
(12) Prove that sin 75° − sin 15° = cos 105° + cos 15°. Solution
(13) Show that tan 75° + cot 75° = 4. Solution
(14) Prove that cos(A + B) cosC−cos(B + C)cosA = sinB sin(C−A). Solution
(15) Prove that sin(n + 1)θ sin(n − 1)θ + cos(n + 1)θ cos(n − 1)θ = cos2θ, n ∈ Z. Solution
(16) If x cos θ = y cos (θ + 2π/3) = z cos (θ + 4π/3), find the value of xy + yz + zx. Solution
(17) Prove that
(i) sin(A + B) sin(A − B) = sin2 A − sin2 B Solution
(ii) cos(A + B) cos(A − B) = cos2 A − sin2 B = cos2 B − sin2 A
(iii) sin2(A + B) − sin2(A − B) = sin2Asin 2B Solution
(iv) cos 8θ cos 2θ = cos2 5θ − sin2 3θ Solution
18. Show that cos2 A + cos2 B − 2 cosAcosB cos(A + B) = sin2(A + B). Solution
19. If cos(α − β) + cos(β − γ) + cos(γ − α) = −3/2, then prove that cos α + cos β + cos γ = sinα + sinβ + sinγ = 0. Solution
(20) Show that
(i) tan(45° + A) = (1 + tanA)/(1 − tanA) Solution
(ii) tan(45° − A) = (1 + tanA)/(1 − tanA) Solution
(21) Prove that cot(A + B) = (cotAcotB − 1)/(cotA + cotB) Solution
22. If tan x = n/(n + 1) and tan y = 1/(2n + 1) , find tan(x + y). Solution
(23) Prove that tan (π/4 + θ) tan (3π/4 + θ) = −1. Solution
(24) Find the values of tan(α+β), given that cot α = 1/2, α ∈ (π, 3π/2) and sec β = −5/3, β ∈ (π/2, π). Solution
(25) If θ + φ = α and tan θ = k tan φ, then prove that sin(θ − φ) = (k − 1)/(k + 1) sin α. Solution
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