using a double - angle formula in exercises 15 - 20, use a double - angle formula to rewrite the…

Explanation:

Step1: Recall double - angle formula

The double - angle formula for sine is $\sin(2x)=2\sin x\cos x$.

Step2: Rewrite 15

For $6\sin x\cos x$, since $\sin(2x) = 2\sin x\cos x$, then $6\sin x\cos x=3\times(2\sin x\cos x)=3\sin(2x)$.

Step3: Rewrite 16

For $\sin x\cos x$, using $\sin(2x)=2\sin x\cos x$, we get $\sin x\cos x=\frac{1}{2}\sin(2x)$.

Step4: Recall double - angle formula for cosine

The double - angle formula for cosine is $\cos(2x)=2\cos^{2}x - 1=1 - 2\sin^{2}x$.

Step5: Rewrite 17

For $6\cos^{2}x-3$, factor out 3: $3(2\cos^{2}x - 1)$. Since $\cos(2x)=2\cos^{2}x - 1$, then $6\cos^{2}x-3 = 3\cos(2x)$.

Step6: Rewrite 18

For $\cos^{2}x-\frac{1}{2}$, rewrite it as $\frac{1}{2}(2\cos^{2}x - 1)$. Since $\cos(2x)=2\cos^{2}x - 1$, then $\cos^{2}x-\frac{1}{2}=\frac{1}{2}\cos(2x)$.

Step7: Rewrite 19

For $4 - 8\sin^{2}x$, factor out 4: $4(1 - 2\sin^{2}x)$. Since $\cos(2x)=1 - 2\sin^{2}x$, then $4 - 8\sin^{2}x=4\cos(2x)$.

Step8: Rewrite 20

For $10\sin^{2}x-5$, factor out 5: $5(2\sin^{2}x - 1)=-5(1 - 2\sin^{2}x)$. Since $\cos(2x)=1 - 2\sin^{2}x$, then $10\sin^{2}x-5=-5\cos(2x)$.

Answer:

15. $3\sin(2x)$
16. $\frac{1}{2}\sin(2x)$
17. $3\cos(2x)$
18. $\frac{1}{2}\cos(2x)$
19. $4\cos(2x)$
20. $-5\cos(2x)$