Question

if 7A = 180 then show that cosA×cos2A×cos3A

Answer 1

the question asked is not ended.the correct question is cosA × cos2A × cos3A =1/8

For this solution:

given: 7A = 180.........(1)
LHS of the given equation is:
cosAcos2Acos3A=cosAcos2Acos(180−4A) [since 3A=180−4A]=cosAcos2A*[−cos4A]=−cosAcos2Acos4A=−12sinA(2sinAcosA)cos2Acos4A [multiplying and dividing by 2sinA=−12sinA*sin2Acos2Acos4A=−14sinA(2sin2Acos2A)cos4A=−14sinAsin4Acos4A=−18sinA(2sin4Aco4A)
=−18sinA*sin8A=−18sinA*sin(7A+A)=−18sinA*sin(180+A)=−18sinA*(−sinA)=18
= RHS

hope this helps you

Answer 2

Answer:Since 7A=180,

sin(7A- x)= sin x

Step-by-step explanation:

LHS=cos A cos 2A cos 3A

Multiplying and dividing by 2 sin A:- =(2 sinA cosA cos 2A cos 3A)÷2 sin A

=(sin 2A cos 2A cos 3A) ÷ 2 sin A

=(2 sin 2A cos 2A cos 3A)÷ 2×2 sin A

=(sin 4A cos 3A)÷4 sin A

=(2 sin 3A cos 3A)÷2×4 sinA

=sin 6A÷8 sin A

=sin A÷ 8 sin A

=1÷8

=RHS

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