If cosx = -2/3 and x is in quadrant II, find sin2x and cos2x.

Solution:

cos2x

= cos²x - sin²x
= cos²x - (1 - cos²x)
= 2cos²x - 1
= 2(-2/3)² - 1
= 8/9 - 1
= -1/9

Now to use the formula sin2x, we need to find sinx first.

sinx

=
=
= It is positive since sinx is positive in quadrant II.

Therefore we get

sin2x

= 2 sinxcosx
= 2(/3)(-2/3)
= -4/9

Example 2:

Find the exact value of cos15^o

Solution:

Using the half-angle formula for cosine with x = 15^o, we get:

cos²15^o

= (1 + cos2(15^o))/2
= (1 + cos30^o)/2
= (1 + /2)/2
= (2 + )/4

Therefore by taking the square root of each side (since 15^o is in the first quadrant, we choose the positive square root) we get

cos15^o

=

=

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