Hi guys, just wanting to check I'm right in my method here as I don't have the mark scheme for this practice paper to hand.
Question 8(a) of June 2006 C3 paper:
Given that cos A= 3/4, where 270°< A <360°, find the exact value of sin 2A
I know double angle formula: sin2A = 2sinAcosA
My working was like this:
If cosA = 3/4. Pythagoras gives us a triangle of √7, 3, 4
Therefore sinA = √7/4
Putting these into sin2A = 2sinAcosA gives: 2(√7/4)(3/4)
This gives us the value of sin2A as: (3√7/8)
Have I gone about this the right way or is it wrong to assume that I can use pythagoras?
Question 8(a) of June 2006 C3 paper:
Given that cos A= 3/4, where 270°< A <360°, find the exact value of sin 2A
I know double angle formula: sin2A = 2sinAcosA
My working was like this:
If cosA = 3/4. Pythagoras gives us a triangle of √7, 3, 4
Therefore sinA = √7/4
Putting these into sin2A = 2sinAcosA gives: 2(√7/4)(3/4)
This gives us the value of sin2A as: (3√7/8)
Have I gone about this the right way or is it wrong to assume that I can use pythagoras?