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Answer to Question #43686 in Trigonometry for yuba raj
Question #43686
prove that sina.cos2a = 1/4 sin4a.sec a
Expert's answer
We can write 1/4 sin(4a)sec(a) as:
1/4 sin(4a)sec(a)=1/4*2sin(2a)*cos(2a)*1/(cos(a))=1/2*2sin(a)*cos(a)*cos(2a)*1/(cos(a))=sin(a)*cos(2a).
Here we use the next formulas: sin(2a)=2*sin(a)*cos(a);
sec(a)=1/(cos(a)).
1/4 sin(4a)sec(a)=1/4*2sin(2a)*cos(2a)*1/(cos(a))=1/2*2sin(a)*cos(a)*cos(2a)*1/(cos(a))=sin(a)*cos(2a).
Here we use the next formulas: sin(2a)=2*sin(a)*cos(a);
sec(a)=1/(cos(a)).
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prove that sin^3a+ cos^3a/sin a + cos a = 1- 1/2 sin2a
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