COMPOUND ANGLE PRACTICE QUESTIONS FOR GRADE 11

Compound Angle Practice Questions for Grade 11 :

Here we are going to see some practice questions on based on compound angles.

Find the Value of Sin x Plus y When Sin x and Cos y are Given - Example problem

(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).

Solution

(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).

Solution

(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

Solution

(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

After having gone through the stuff given above, we hope that the students would have understood "Compound Angle Practice Questions for Grade 11". 

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