Verify as an identity: cos(90°+x) = -sinx I'm not sure how to go about this question

1 answer:

3 0

Cos( a + b ) = cosa×cosb - sina×sinb ;
a = 90 and b = x ;
But, cos90 = 0 and sin90 = 1;
cos( 90 + x ) = 0 × cosx - 1 × sinx = - sinx ;

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8 0

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mihalych1998 [28]

This is a doozy so pay attention. First thing you have to recognize is that is a trig identity, the sum identity for cosine to be exact. That formula is

$cos(\alpha +\beta )=cos\alpha cos\beta -sin\alpha sin\beta$ .

So that means you need to find the cosine of alpha and the sin of beta. We are given the sin of alpha being 4/5 in the first quadrant. If you set up the sin ratio which is side opposite over hypotenuse of a right triangle, you put the 4 on the side opposite the reference angle, alpha, and the hypotenuse of 5 on the terminal ray and then you have to find the missing side of the right triangle you created. Using Pythagorean's theorem, you find that the missing side, which is the adjacent side to alpha, is 3. Now you can find the cosine of alpha as well, since cosine is the side adjacent, 3, over the hypotenuse, 5. So far wwe have

$sin\alpha =\frac{4}{5}$ and

$cos\alpha =\frac{3}{5}$ for that first angle. Now moving on to the second angle. The cosine of beta is side adjacent, 5, over the hypotenuse, 13, and we are missing the side opposite the reference angle beta. Using Pythagorean's theorem again to find the side opposite, we have that that side measures 12. Now we can find the sine of beta using that opposite side, 12, over the hypotenuse, 13. What we have now is