Let α and β be first quadrant angles with cos ( α ) = √ 11 4 and sin ( β ) = √ 3 8 . Find sin ( α − β ) .

Question

I need help with this question. I'm aware of the formula that I need to use. What I'm having trouble is working with the radicals. Can someone show me the process to work with them? Thank you

Answer 1

Using a and b instead of α and β. Easier to type.

Not clear from above, but I'm assuming cos(a) = √11 / 4 and sin(b) = √3 / 8. Where 4 and 8 are the hypotenuses of each of the two triangles. √11 is the adjacent side of one and √3 is the opposite side of the other.

That would make sin(a) = √(16 - 11)/4 = √5/4 and cos(b) = √(64-3)/8 = √61 / 8.

sin(a-b) = sin(a)cos(b) - sin(b)cos(a)

= √5/4√61/8 - √3/8√11/4 = (√(5•61) - √(3•11))/(4•8) = (√305 - √33)/32

I don't think this can be simplified.

Answer 2

Your values for cos and sin are not possible as both functions have to be between -1 and 1

So you need both cos(β) and sin(α) to plug into the sin(α-β) equation.

I'll do cos(α) and the first term of sin(α-β):

sin(α) = sqrt( 1 - cos²(α))

When you multiply sin(α)cos(β), just multiply the parts under the sqrt sign:

Making up numbers : sqrt(.1)* sqrt(.2) = sqrt(.02)

Good luck.