Rewrite the equation in terms of t = sin x t=\sin x t = sin x
Given
cos 2 x + a sin x = 2 a − 7. \cos 2x + a\sin x = 2a-7. cos 2 x + a sin x = 2 a − 7. Using
cos 2 x = 1 − 2 sin 2 x , \cos 2x = 1-2\sin^2 x, cos 2 x = 1 − 2 sin 2 x , we get
1 − 2 sin 2 x + a sin x = 2 a − 7. 1-2\sin^2 x + a\sin x = 2a-7. 1 − 2 sin 2 x + a sin x = 2 a − 7. Let t = sin x t=\sin x t = sin x t ∈ [ − 1 , 1 ] t\in[-1,1] t ∈ [ − 1 , 1 ]
1 − 2 t 2 + a t = 2 a − 7. 1-2t^2+at=2a-7. 1 − 2 t 2 + a t = 2 a − 7. Rearranging,
2 t 2 − a t + ( 2 a − 8 ) = 0. 2t^2-at+(2a-8)=0. 2 t 2 − a t + ( 2 a − 8 ) = 0. Solve for a a a t t t
a t − 2 a = 2 t 2 − 8 at-2a = 2t^2-8 a t − 2 a = 2 t 2 − 8 a ( t − 2 ) = 2 ( t 2 − 4 ) = 2 ( t − 2 ) ( t + 2 ) . a(t-2)=2(t^2-4)=2(t-2)(t+2). a ( t − 2 ) = 2 ( t 2 − 4 ) = 2 ( t − 2 ) ( t + 2 ) . Since t ∈ [ − 1 , 1 ] t\in[-1,1] t ∈ [ − 1 , 1 ] t ≠ 2 t\neq 2 t = 2 t − 2 t-2 t − 2
a = 2 ( t + 2 ) = 2 t + 4. a=2(t+2)=2t+4. a = 2 ( t + 2 ) = 2 t + 4. Now t ∈ [ − 1 , 1 ] t\in[-1,1] t ∈ [ − 1 , 1 ]
a = 2 t + 4 ∈ [ 2 ( − 1 ) + 4 , 2 ( 1 ) + 4 ] = [ 2 , 6 ] . a=2t+4 \in [2(-1)+4,\,2(1)+4]=[2,6]. a = 2 t + 4 ∈ [ 2 ( − 1 ) + 4 , 2 ( 1 ) + 4 ] = [ 2 , 6 ] . Hence,
p = 2 , q = 6. p=2,\quad q=6. p = 2 , q = 6.
Evaluate r r r
Given
r = tan 9 ∘ − tan 27 ∘ − 1 cot 63 ∘ + tan 81 ∘ . r=\tan 9^\circ-\tan 27^\circ-\frac{1}{\cot 63^\circ}+\tan 81^\circ. r = tan 9 ∘ − tan 2 7 ∘ − cot 6 3 ∘ 1 + tan 8 1 ∘ . Now,
1 cot 63 ∘ = tan 63 ∘ . \frac{1}{\cot 63^\circ}=\tan 63^\circ. cot 6 3 ∘ 1 = tan 6 3 ∘ . So
r = tan 9 ∘ − tan 27 ∘ − tan 63 ∘ + tan 81 ∘ . r=\tan 9^\circ-\tan 27^\circ-\tan 63^\circ+\tan 81^\circ. r = tan 9 ∘ − tan 2 7 ∘ − tan 6 3 ∘ + tan 8 1 ∘ . Use
tan ( 90 ∘ − θ ) = cot θ . \tan(90^\circ-\theta)=\cot\theta. tan ( 9 0 ∘ − θ ) = cot θ . Thus
tan 81 ∘ = cot 9 ∘ , tan 63 ∘ = cot 27 ∘ . \tan 81^\circ=\cot 9^\circ,\qquad \tan 63^\circ=\cot 27^\circ. tan 8 1 ∘ = cot 9 ∘ , tan 6 3 ∘ = cot 2 7 ∘ . Hence
r = ( tan 9 ∘ + cot 9 ∘ ) − ( tan 27 ∘ + cot 27 ∘ ) . r=(\tan 9^\circ+\cot 9^\circ)-(\tan 27^\circ+\cot 27^\circ). r = ( tan 9 ∘ + cot 9 ∘ ) − ( tan 2 7 ∘ + cot 2 7 ∘ ) . Now,
tan θ + cot θ = sin 2 θ + cos 2 θ sin θ cos θ = 1 sin θ cos θ = 2 sin 2 θ . \tan\theta+\cot\theta=\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}
=\frac{1}{\sin\theta\cos\theta}
=\frac{2}{\sin 2\theta}. tan θ + cot θ = sin θ cos θ sin 2 θ + cos 2 θ = sin θ cos θ 1 = sin 2 θ 2 . Therefore,
tan 9 ∘ + cot 9 ∘ = 2 sin 18 ∘ , \tan 9^\circ+\cot 9^\circ=\frac{2}{\sin 18^\circ}, tan 9 ∘ + cot 9 ∘ = sin 1 8 ∘ 2 , tan 27 ∘ + cot 27 ∘ = 2 sin 54 ∘ . \tan 27^\circ+\cot 27^\circ=\frac{2}{\sin 54^\circ}. tan 2 7 ∘ + cot 2 7 ∘ = sin 5 4 ∘ 2 . So
r = 2 sin 18 ∘ − 2 sin 54 ∘ . r=\frac{2}{\sin 18^\circ}-\frac{2}{\sin 54^\circ}. r = sin 1 8 ∘ 2 − sin 5 4 ∘ 2 . Use the known values
sin 18 ∘ = 5 − 1 4 , \sin 18^\circ=\frac{\sqrt5-1}{4}, sin 1 8 ∘ = 4 5 − 1 , sin 54 ∘ = cos 36 ∘ = 5 + 1 4 . \sin 54^\circ=\cos 36^\circ=\frac{\sqrt5+1}{4}. sin 5 4 ∘ = cos 3 6 ∘ = 4 5 + 1 . Then
r = 8 5 − 1 − 8 5 + 1 . r=\frac{8}{\sqrt5-1}-\frac{8}{\sqrt5+1}. r = 5 − 1 8 − 5 + 1 8 . Combine:
r = 8 ( ( 5 + 1 ) − ( 5 − 1 ) ( 5 − 1 ) ( 5 + 1 ) ) = 8 ( 2 5 − 1 ) = 8 ⋅ 1 2 = 4. r=8\left(\frac{(\sqrt5+1)- (\sqrt5-1)}{(\sqrt5-1)(\sqrt5+1)}\right)
=8\left(\frac{2}{5-1}\right)=8\cdot \frac12=4. r = 8 ( ( 5 − 1 ) ( 5 + 1 ) ( 5 + 1 ) − ( 5 − 1 ) ) = 8 ( 5 − 1 2 ) = 8 ⋅ 2 1 = 4. Thus,
r = 4. r=4. r = 4.
Compute p q r pqr pq r
p q r = 2 ⋅ 6 ⋅ 4 = 48. pqr = 2\cdot 6\cdot 4 = 48. pq r = 2 ⋅ 6 ⋅ 4 = 48. Therefore, the required integer is
48 . \boxed{48}. 48 .
Comparison with stored answer
Stored correct answer = 48 48 48
Our derived answer also is 48 48 48