Answer:
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∘
Answer:
Step-by-step explanation:
3
6. I think it's b but I could be wrong hope I helped.
