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Use the power-reducing formulas to prove
We will work on simplifying the left side of the equation:
Use the power-reducing formulas to prove that 10 cos4x=154+5 cos(2x)+54 cos(4x).
10cos4x=10(cos2x)2=10[1+cos(2x)2]2Substitute reduction formula for cos2x.=104[1+2cos(2x)+cos2(2x)]=104+102cos(2x)+104(1+cos2(2x)2)Substitute reduction formula for cos2x.=104+102cos(2x)+108+108cos(4x)=308+5cos(2x)+108cos(4x)=154+5cos(2x)+54cos(4x)
The next set of identities is the set of half-angle formulas , which can be derived from the reduction formulas and we can use when we have an angle that is half the size of a special angle. If we replace θ with α2, the half-angle formula for sine is found by simplifying the equation and solving for sin(α2). Note that the half-angle formulas are preceded by a ± sign. This does not mean that both the positive and negative expressions are valid. Rather, it depends on the quadrant in which α2 terminates.
The half-angle formula for sine is derived as follows:
To derive the half-angle formula for cosine, we have
For the tangent identity, we have
The half-angle formulas are as follows:
Find sin(15°) using a half-angle formula.
Since 15°=30°2, we use the half-angle formula for sine:
Remember that we can check the answer with a graphing calculator.
Given the tangent of an angle and the quadrant in which the angle lies, find the exact values of trigonometric functions of half of the angle.
Given that tan α=815 and α lies in quadrant III, find the exact value of the following:
Using the given information, we can draw the triangle shown in [link] . Using the Pythagorean Theorem, we find the hypotenuse to be 17. Therefore, we can calculate sin α=−817 and cos α=−1517.
To find sin α2, we begin by writing the half-angle formula for sine. Then we substitute the value of the cosine we found from the triangle in [link] and simplify.
We choose the positive value of sin α2 because the angle terminates in quadrant II and sine is positive in quadrant II.
We choose the negative value of cos α2 because the angle is in quadrant II because cosine is negative in quadrant II.
We choose the negative value of tan α2 because α2 lies in quadrant II, and tangent is negative in quadrant II.
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