Question

prove that√2sin10+√3cos35=sin55+2cos65​

Answer 1

Answer:

As Deepak wrote it's easier to start with the right hand side, but you could go ahead with your approach too, as shown by Arian.

2–√sin10∘+3–√cos35∘=sin55∘+2cos65∘(1)

(1)2sin⁡10∘+3cos⁡35∘=sin⁡55∘+2cos⁡65∘

Here is a variant that uses the cosine addition formula for 65∘=35∘+30∘65∘=35∘+30∘, the sine difference formula for 10∘=45∘−35∘10∘=45∘−35∘ and the sine/cosine complement formula.

Use the complement formula sinθ=cos(90∘−θ)sin⁡θ=cos⁡(90∘−θ) for θ=55∘θ=55∘, the addition formula cos(a+b)=cosacosb−sinasinb,cos⁡(a+b)=cos⁡acos⁡b−sin⁡asin⁡b, and the special values sin30∘=12sin⁡30∘=12, cos30∘=3√2cos⁡30∘=32 to rewrite the right hand side of (1)(1) as [hover your mouse over the grey to see]

Answer 2

Step-by-step explanation:

Deepak wrote it's easier to start with the right hand side, but you could go ahead with your approach too, as shown by Arian.

2–√sin10∘+3–√cos35∘=sin55∘+2cos65∘(1)

Here is a variant that uses the cosine addition formula for 65∘=35∘+30∘, the sine difference formula for 10∘=45∘−35∘ and the sine/cosine complement formula.

Use the complement formula sinθ=cos(90∘−θ) for θ=55∘, the addition formula cos(a+b)=cosacosb−sinasinb, and the special values sin30∘=12, cos30∘=3√2 to rewrite the right hand side of (1) as [hover your mouse over the grey to see]