T. (a) Prove that $$ \sin 2 x+2 \sin ^{2}\left(45^{\circ}-x\right)=1 $$

Question

UpStudy AI · Accepted Answer

Below is a step‐by‐step proof that

$$
\sin 2x + 2 \sin^2\left(45^\circ - x\right) = 1.
$$

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### Step 1. Use the Double-Angle Identity for Sine Squared

Recall the identity for the sine squared of an angle:

$$
\sin^2 \theta = \frac{1 - \cos 2\theta}{2}.
$$

Let $$\theta = 45^\circ - x$$. Then we have:

$$
\sin^2\left(45^\circ - x\right) = \frac{1 - \cos\left(2(45^\circ - x)\right)}{2}.
$$

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### Step 2. Substitute the Expression in the Given Equation

Multiply the expression by 2 (as given in the original equation):

$$
2\sin^2\left(45^\circ - x\right) = 2 \cdot \frac{1 - \cos\left(90^\circ - 2x\right)}{2} = 1 - \cos\left(90^\circ - 2x\right).
$$

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### Step 3. Relate $$\cos(90^\circ - 2x)$$ to $$\sin 2x$$

Use the complementary angle identity for cosine:

$$
\cos\left(90^\circ - 2x\right) = \sin 2x.
$$

Substitute this into our expression:

$$
2\sin^2\left(45^\circ - x\right) = 1 - \sin 2x.
$$

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### Step 4. Combine the Terms

Now, add $$\sin 2x$$ to both sides of the equation:

$$
\sin 2x + 2\sin^2\left(45^\circ - x\right) = \sin 2x + \left(1 - \sin 2x\right) = 1.
$$

Thus, we have shown that

$$
\sin 2x + 2 \sin^2\left(45^\circ - x\right) = 1.
$$

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### Conclusion

Using trigonometric identities and algebraic manipulation, we proved the required equality.
$$ \sin 2x + 2 \sin^2\left(45^\circ - x\right) = 1 $$

T. (a) Prove that \( \sin 2 x+2 \sin ^{2}\left(45^{\circ}-x\right)=1 \)