cos2β if sinβ = 4/5 and β is in the first quadrant
Im having a hard time undersatanding this.
recall your double-angle formula
cos2β = 1 - 2sin^2β = 1 - 2(4/5)^2 = 1 - 2*16/25 = -7/25
This makes sense, since β > π/4, so 2β > π/2, in QII
You will do well to do lots of these problems, to make sure you know the double-angle formulas by heart. They come up all the time.
Thank you very very much!!
To find cos(2β), we'll need to use the double-angle identity of cosine. The double-angle identity for cosine states that:
cos(2β) = cos²(β) - sin²(β)
Given that sin(β) = 4/5 and β is in the first quadrant, we can use this information to find cos²(β) and sin²(β), and then substitute them into the double-angle identity to find cos(2β).
Let's start by finding sin²(β). We know that sin(β) = 4/5. Since sin²(β) = (sin(β))², we can square both sides of the equation:
sin²(β) = (4/5)²
sin²(β) = 16/25
Next, we can find cos²(β) using the Pythagorean identity for trigonometric functions:
cos²(β) = 1 - sin²(β)
cos²(β) = 1 - 16/25
cos²(β) = 9/25
Now we have cos²(β) and sin²(β), so we can substitute them into the double-angle identity of cosine:
cos(2β) = cos²(β) - sin²(β)
cos(2β) = 9/25 - 16/25
cos(2β) = -7/25
Therefore, if sin(β) = 4/5 and β is in the first quadrant, cos(2β) = -7/25.