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4 years ago

Evaluate Cos^-1(sin 330 degree)

1 answer:

5 0

$330^\circ\equiv-30^\circ\mod360^\circ$

so

$\sin330^\circ=\sin(-30^\circ)=-\sin30^\circ=-\dfrac12$

Finally,

$\cos^{-1}\left(-\dfrac12\right)=\dfrac{2\pi}3\text{ rad}=120^\circ$

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3 years ago

Does anyone know how to solve this?

Charra [1.4K]

Answer:

$\frac{\sqrt{2}}{2}$

Step-by-step explanation:

This almost looks like the left hand side of the following identity:

$\sin(A)\cos(B)-\sin(B)\cos(A)=\sin(A-B)$ .

Here are similar identities in the same category as the above:

$\sin(A)\cos(B)+\sin(B)\cos(A)=\sin(A+B)$

$\cos(A)\cos(B)-\sin(A)\sin(B)=\cos(A+B)$

$\cos(A)\cos(B)+\sin(A)\sin(B)=\cos(A-B)$

Things to notice: 90-76=14. and 90-59=31.

This means we will possibly want to use the following co-function identities:

$\cos(90-A)=\sin(A)$

$\sin(90-A)=\cos(A)$

So let's begin:

$\sin(76)\cos(31)-\sin(14)\cos(59)$

Applying the co-function identities:

$\cos(14)\cos(31)-\sin(14)\sin(31)$

Applying one of the difference identities above with cosine:

$\cos(31+14)$

$\cos(45)$

45 is a special angle so $\cos(45)$ is something you find off most unit circles in any trigonometry class.

$\cos(45)=\frac{\sqrt{2}}{2}$

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4 years ago