Question

show that cos38°cos52°-sin52°=cos 90°​

Answer 1

Answer:

I am pretty sure you missed out a term and the original question is supposed to be show that cos38°cos52° - sin52°sin38° = cos 90°

Solving LHS:

cos38°cos52° - sin52°sin38°

= cos (90 - 52)°cos(90-38)° - sin52°sin38°

= sin52°sin38° - sin52°sin38°

=0

=cos 90°