Prove: (3cosx + 4sinx)^2 + (4cosx - 3sinx)^2=5 ...this is all under the square root sign.
I get 9cos^2x + 16sin^2x +16cos^2x + 9sin^2x
now i know that sin^2 equals 1-cos^2....so would that be 9cos^2x + 17-cos^2x+16cos^2+ 10cos^2x? if not please help..and if this is correct..what would come after this step?
continuing from your left side you had:
9cos^2x + 16sin^2x +16cos^2x + 9sin^2x
this is ok so far.
group this into 9(sin^2 x + cos^2 x) + 16(sin^2 x + cos^2 x)
which is 9(1) + 16(1), because sin^2 x + cos^2 x = 1
So the left side is 25
You said that there was a square root over everything?
so...., how about that!
To prove the given statement, let's simplify the expression on the left side step by step:
Starting with (3cosx + 4sinx)^2 + (4cosx - 3sinx)^2 under the square root:
= (√[(3cosx + 4sinx)^2 + (4cosx - 3sinx)^2])
Squaring both terms inside the bracket:
= (√(9cos^2x + 24cosxsinx + 16sin^2x + 16cos^2x - 24cosxsinx + 9sin^2x))
Combining like terms:
= (√(25cos^2x + 25sin^2x))
Using the trigonometric identity cos^2x + sin^2x = 1:
= (√(25(1)))
Simplifying further:
= (√(25))
Taking the square root of 25:
= 5
So, the left side simplifies to 5.