Prove: (3cosx + 4sinx)^2 + (4cosx - 3sinx)^2=5 ...this is all under the square root sign.
this is what i did please check.
i multipied both binomials. but i get cos square x..and sincosx ..how do i combine them..please help!
In this case, after multiplying binomials, there will be 24sincos and -24sincos terms that cancel. That will leave you with sin^2 and cos^2 terms. Write sin^2 as 1-cos^2 and solve for cosx
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?
To solve the given expression, let's start from where you left off:
(3cosx + 4sinx)^2 + (4cosx - 3sinx)^2 = 9cos^2x + 16sin^2x + 16cos^2x + 9sin^2x
Now, we can simplify further using the identities sin^2x = 1 - cos^2x and cos^2x = 1 - sin^2x:
= 9cos^2x + 16(1 - cos^2x) + 16cos^2x + 9(1 - cos^2x)
= 9cos^2x + 16 - 16cos^2x + 16cos^2x + 9 - 9cos^2x
The cos^2x terms in the equation cancel out:
= 9 + 16 + 9
= 34
Therefore, the result is 34.
To summarize the steps:
1. Multiply the binomials to expand the expression.
2. Apply the identities sin^2x = 1 - cos^2x and cos^2x = 1 - sin^2x to simplify the expression.
3. Simplify the equation further by combining like terms.
4. The final result is 34.
Remember to be careful when applying trigonometric identities and double-check your calculations.