Q.Integrate the following function: $\sin^2 (2x + 5)$

We are given $A+B+C = 4S$. The expression to simplify is

The trick is to notice that each term involves $2S$ minus one of $A, B, C$. Using the given relation, we can express $2S$ in terms of the half-angles, making the sum-to-product identities very clean.

1. Rewrite each angle using $S$ From $A+B+C = 4S$, we have $S = \frac{A+B+C}{4}$. Then

Similarly,

And $2S = \frac{A+B+C}{2}$.

2. Pair the first two sines and apply sum-to-product Consider $\sin(2S-A) + \sin(2S-B)$. Using

with $X = \frac{-A+B+C}{2}$ and $Y = \frac{A-B+C}{2}$:

So

3. Add the third sine and subtract $\sin 2S$ Now we have

Write $\sin(2S-C) = \sin\frac{A+B-C}{2}$ and $\sin 2S = \sin\frac{A+B+C}{2}$.

Pair these two:

Use $\sin P - \sin Q = 2\cos\frac{P+Q}{2}\sin\frac{P-Q}{2}$:

So

4. Combine everything The whole expression becomes

Use the identity $\cos u - \cos v = -2\sin\frac{u+v}{2}\sin\frac{u-v}{2}$:

Simplify the arguments:

So

5. Final simplification Substituting back:

Since $A+B+C=2S$, we have $2S-A-B=C$ and $2S-C=A+B$.

Expand each product with $\sin P\cos Q=\tfrac12\big[\sin(P+Q)+\sin(P-Q)\big]$:

Subtracting,

Now $\sin(B-A)-\sin(A+B)=-2\cos B\sin A$, and with the angle-closure $\sin C=\sin(A+B)$ this becomes

The key idea is to rewrite the left‑hand side using the half‑angle formulas for $\cos^2\frac{C}{2}$ and $\sin^2\frac{C}{2}$ in terms of $\cos C$. That will turn the equation into a relation between the sides $a$, $b$, $c$ and the angle $C$. Once we see that relation, we can deduce the shape of the triangle and then find $\cos A$.

1. Recall the half‑angle identities For any angle $C$,

These are standard and come from the double‑angle formulas.

2. Substitute into the given equation The equation is

Replace the squares:

3. Multiply through by 2 and expand

Expand each term:

4. Group the constant terms and the $\cos C$ terms First, the constant part (terms without $\cos C$):

The $\cos C$ part:

Simplify the bracket:

So the $\cos C$ part is $-4ab\cos C$.

5. Combine everything The left side becomes

The equation is

Cancel $2a^2+2b^2$ from both sides, leaving

Since $a>0$, $b>0$, we get $\cos C = 0$.

6. Interpret $\cos C = 0$

   In a triangle, $0 < C < \pi$, so $\cos C = 0$ implies $C = 90^\circ$. Hence triangle $ABC$ is right‑angled at $C$.

7. Find $\cos A$ in this right triangle

   In a right triangle with right angle at $C$, the other two angles $A$ and $B$ are acute and complementary: $A + B = 90^\circ$.

   Therefore $\cos A = \sin B$ (since $\cos\theta = \sin(90^\circ - \theta)$).

   Looking at the options:

   (A) $\cos B$ — not generally equal to $\cos A$ unless $A=B$.

   (B) $\sin C = \sin 90^\circ = 1$, not equal to $\cos A$ unless $A=0^\circ$.

   (C) $\sin B$ — yes, exactly.

   (D) $\cos C = 0$, not equal to $\cos A$ unless $A=90^\circ$.

The key insight is that the angles $\frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8}$ are symmetric about $\frac{\pi}{2}$. This symmetry lets us pair terms and use the identity $\sin^4 x + \cos^4 x = 1 - \frac12 \sin^2 2x$, which simplifies the sum dramatically without needing to compute each fourth power separately.

1. Recognize the symmetry Notice that $\frac{5\pi}{8} = \pi - \frac{3\pi}{8}$ and $\frac{7\pi}{8} = \pi - \frac{\pi}{8}$. Since $\cos(\pi - \alpha) = -\cos\alpha$ and $\sin(\pi - \alpha) = \sin\alpha$, their fourth powers are the same: $\cos^4\frac{5\pi}{8} = \cos^4\frac{3\pi}{8}$ and $\sin^4\frac{7\pi}{8} = \sin^4\frac{\pi}{8}$. So the sum becomes:

2. Pair the terms Group them as:

3. Apply the identity For any angle $x$, we have $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac12 \sin^2 2x$. So:

Similarly, for $x = \frac{3\pi}{8}$, note that $2x = \frac{3\pi}{4}$, and $\sin\frac{3\pi}{4} = \frac{\sqrt{2}}{2}$ as well. So:

4. Sum the two pairs

5. Set up the equation for $\theta$ The problem states $2\sin^2\theta = S$, so:

6. Find $\theta$ The acute angle with sine $\frac{\sqrt{3}}{2}$ is $\theta = \frac{\pi}{3}$.

The angles in the sine product run from $\pi/14$ to $13\pi/14$ in steps of $2\pi/14$, covering every odd multiple of $\pi/14$. Using $\sin(\pi-\theta)=\sin\theta$:

• $\sin(13\pi/14) = \sin(\pi/14)$
• $\sin(11\pi/14) = \sin(3\pi/14)$
• $\sin(9\pi/14) = \sin(5\pi/14)$
• $\sin(7\pi/14) = \sin(\pi/2) = 1$

So the seven factors pair into three repeated pairs plus the middle term:

Step-by-step solution

1. Convert sines to cosines using $\sin\theta = \cos(\pi/2-\theta)$:

So $P = \cos\tfrac{3\pi}{7}\cos\tfrac{2\pi}{7}\cos\tfrac{\pi}{7}$.

2. Relate $\cos(3\pi/7)$ to $\cos(4\pi/7)$ using $\cos(\pi-x)=-\cos x$, with $3\pi/7 = \pi - 4\pi/7$:

so

3. Evaluate the classic product $\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}$ using repeated application of $\sin(2\theta)=2\sin\theta\cos\theta$:

(Note: the stem's printed value "$8\pi/7$" for this product is a misprint — a product of three cosines is a pure number, not a multiple of $\pi$; the correct exact value is $-1/8$.)

4. Substitute back: $P = -\left(-\tfrac{1}{8}\right) = \tfrac{1}{8}$.

5. Square to get the full product: $\prod = P^2 = \left(\tfrac{1}{8}\right)^2 = \tfrac{1}{64}$.

We start with the given:

$\sin\alpha + \cos\alpha = m$.

We want $\sin^6\alpha + \cos^6\alpha$.

Concept and intuition:

The expression $\sin^6\alpha + \cos^6\alpha$ is a sum of sixth powers. A classic trick is to rewrite it as $(\sin^2\alpha)^3 + (\cos^2\alpha)^3$, then use the sum of cubes factorization:

$a^3+b^3 = (a+b)^3 - 3ab(a+b)$.

Here $a=\sin^2\alpha$, $b=\cos^2\alpha$, so $a+b=1$ (since $\sin^2\alpha+\cos^2\alpha=1$). That reduces the problem to finding $\sin^2\alpha\cos^2\alpha$, which we can get from $m$ because $m^2 = 1 + 2\sin\alpha\cos\alpha$.

Let’s work it out step by step.

1. Express the sixth power sum using cubes.

Let $u = \sin^2\alpha$, $v = \cos^2\alpha$. Then $u+v=1$, and