Question

a) Show that 2sin B + 4cos(a+ß) sina sinB+ cos 2(a +B) = cos2a.
b) Solve Square root 3 Cosò=square root 2
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Answer 1

Answer:

a=2sin

2

β+4cos(α+β)sinαsinβ+cos2(α+β)

⇒a=2sin

2

β+2cos(α+β)(2sinαsinβ)+cos2(α+β)

⇒a=1−cos2β+2cos(α+β)(cos(α−β)−cos(α+β))+cos2(α+β)

...{∵cos2A=1−2sin

2

A&2sinAsinB=cos(A−B)−cos(A+B)}

⇒a=1−cos2β+(2cos(α+β)cos(α−β))−(2cos

2

(α+β))+cos2(α+β)

⇒a=1−cos2β+(2cos(α+β)cos(α−β))−(2cos

2

(α+β))+cos2(α+β)

⇒a=1−cos2β+(cos2α+cos2β)−(1+cos2(α+β))+cos2(α+β)

...{∵cos2A=2cos

2

A−1&2cosAcosB=cos(A+B)+cos(A−B)}

∴a=cos2α