Sin(θ-30) = cos(θ)
solve for θ.​

Step-by-step explanation:

sin(θ+30∘)=cos50∘

⟹cos(90∘−(θ+30∘))=cos50∘

⟹cos(60∘−θ)=cos50∘

⟹cos(π3−θ)=cos5π18

Writing the general solution as follows

π3−θ=2nπ±5π18

⟹θ=π3−(2nπ±5π18)

Method 2: ,

sin(θ+30∘)=cos50∘

⟹sin(θ+30∘)=sin(90∘−50∘)

⟹sin(θ+30∘)=sin40∘

⟹sin(θ+π6)=sin2π9

Writing the general solution as follows

θ+π6=2nπ+2π9

⟹θ=2nπ+2π9−π6

⟹θ=2nπ+π18

or

θ+π6=(2n+1)π−2π9

⟹θ=2nπ+π−2π9−π6

⟹θ=2nπ+11π18

Hint 1: sin(a)=sin(b) iff a−b=2kπ or a+b=(2k+1)π for some k∈Z.

Hint 2: cos(40∘)=sin(50∘).

Hint:

sinθ=cos(90∘−θ)

cos50∘=sin40∘

can you solve for θ using the above?

0

Knowing the relation between sin(θ) and cos(θ) is quite crucial. One of the major relation is that the sine function and cosine function are fairly similar with 90∘ difference so,

Sin(x+90)=cos(x)

We are given x=50, so

x+90=30+θ

θ=110

or

180−140=40

This is θ+30 so,

θ=10∘

yes because there is exactly one y value for every x value

for function there can't be more than one value for every x value