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Kruka [31]
4 years ago
5

PROVE THAT:cos 20° - sin 20° = ​​ \sqrt{2}sin25°​

Mathematics
1 answer:
Lemur [1.5K]4 years ago
4 0

Answer:

See below.

Step-by-step explanation:

\cos(20)-\sin(20)=\sqrt{2}\sin(25)

First, use the co-function identity:

\sin(90-x)=\cos(x)

We can turn the second term into cosine:

\sin(20)=\sin(90-70)=\cos(70)

Substitute:

\cos(20)-\cos(70)=\sqrt{2}\sin(25)

Now, use the sum to product formulas. We will use the following:

\cos(x)-\cos(y)=-2\sin(\frac{x+y}{2})\sin(\frac{x-y}{2})

Substitute:

\cos(20)-\cos(70)=-2\sin(\frac{20+70}{2})\sin(\frac{20-70}{2})\\\cos(20)-\cos(70)  =-2\sin(45)\sin(-25)\\\cos(20)-\cos(70)=-2(\frac{\sqrt{2}}{2})\sin(-25)\\ \cos(20)-\cos(70)=-\sqrt{2}\sin(-25)

Use the even-odd identity:

\sin(-x)=-\sin(x)

Therefore:

\cos(20)-\cos(70)=-\sqrt{2}\sin(-25)\\\cos(20)-\cos(70)=-\sqrt{2}\cdot-\sin(25)\\\cos(20)-\cos(70)=\sqrt{2}\sin(25)

Replace the second term with the original term:

\cos(20)-\sin(20)=\sqrt{2}\sin(25)

Proof complete.

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