# 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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## Comments

Assignment Expert27.06.14, 20:18Dear yuba raj,

please write your question in the "Questions and Answers" on the site (www.AssignmentExpert.com). Thank You for your question.

yuba raj27.06.14, 05:38prove that sin^3a+ cos^3a/sin a + cos a = 1- 1/2 sin2a

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