Question

Please help with my math problem! Thank you :)

Answer 1

Answer:

cosΘ = $\frac{11}{12}$

Step-by-step explanation:

Using the trigonometric identity

sin²x + cos²x = 1 ⇒ cosx = ± $\sqrt{1-sin^2x}$

Given

sinΘ = $\frac{\sqrt{23} }{12}$, then

cosΘ = ± $\sqrt{1-(\frac{\sqrt{23} }{12} }$)² = ± $\sqrt{1-\frac{23}{144} }$ = ± $\sqrt{\frac{121}{144} }$ = ± $\frac{11}{12}$

Since Θ is in first quadrant then cosΘ > 0

cosΘ = $\frac{11}{12}$

Answer 2

Answer:

cos Ф = 11/12

Step-by-step explanation:

Given that:

sinФ = $\sqrt{23}/12$

Putting in Pythagoras theorem:

($\sqrt{23}/12$)^2 + cos^2 Ф = 1

23/144 + cos^2 Ф = 1

Subtracting 23/144 on both sides:

cos^2 Ф = 1 - 23/144

By taking LCM

cos^2 Ф = (144 - 23)/144

cos^2 Ф = 121/144

Taking square root on both sides we get:

cos Ф = 11/12

I hope it will help you!