A few quick answers needed (1 Viewer)

[<Stretch>]

If sinx = 3/5, find sin2x and cos 2x where
a) 0 < x < pi/2
b) pi/2 < x < pi

and 2 induction Q's Prove:
a) 7^n -3^n is divisible by 4 for all n >/ 1
b) 1/2! + 2/3! + 3/4! +.....+ ((n+1)! - 1) / (n+1)!

Thanks to anyone with help.

 

[_ShiFTy_]

sin2x = 2sinxcosx
cos2x = (cosx)^2 - (sinx)^2

 

[Trev]

7ⁿ-3ⁿ divisable by 4 for all n>=1.
Test for n=1.
I forget the structure of it, but this is what you do:
4P=7ⁿ-3ⁿ
4Q=7ⁿ⁺¹-3ⁿ⁺¹
3ⁿ=7ⁿ-4P
So,
4Q=7ⁿ⁺¹-3.3ⁿ
4Q=7ⁿ⁺¹-3(7ⁿ-4P)
4Q=7.7ⁿ-3.7ⁿ+12P
4Q=4.7ⁿ+12P
4Q=4(7ⁿ+3P)
Blah, therefore divisable by 4 for all n>=1.

 

[Riviet]

If sinx = 3/5, find sin2x and cos 2x where
a) 0 < x < pi/2
b) pi/2 < x < pi

Draw a right triangle ABC where C is the vertex with the right angle, label angle BAC as x^o, so sinx=BC/AB=3/5, which is given in question. By inspection that it's a 3-4-5 triangle, side AC=4, so cosx=4/5

a)Since the restriction for is x is 0 < x < pi/2, all trig functions are positive, so we don't need to worry about any signs.
Now sin2x=2sinxcosx
substitute our values of sinx=3/5 and cosx=4/5,
=2 x 3/5 x 4/5
=24/25

So cos2x=cos²x-sin²x
=(4/5)²-(3/5)²
=16/25 - 9/25
=7/25

b) pi/2 < x < pi implies the second quadrant, where sine is positive and cosine is negative, so our values are now sinx=3/5 and cosx=-4/5, Just do the same as part a) but with these values of sinx and cosx.
Hope that helps.