prove that: sin²(3x) + cos²(3x) + sin²(x) - cos²(x) = 0
come on you know that
sin²(u) + cos²(u) = 1
so, the equation reduces to
sin²(3x) + cos²(3x) - (sin²(x) + cos²(x)) = 0
1 - 1 = 0
well, easy
sin^2 (3x)+cos^2(3x)=1
that leaves on the left
1-cos^2 x + sin^2 x
but 1-cos^2 x=sin^2 x
so, now you are left with
sin^2 x + sin^2 x which is NOT zero. The equation is false.
sin²(3x) + cos²(3x) = 1
1 + sin²(x) - cos²(x) = 2sin²(x)
not zero
I suspect you have a typo. A true assertion is
sin²(3x) + cos²(3x) - sin²(x) - cos²(x) = 0
thank you
then how do you solve : sin²(3x) + cos²(3x) - sin²(x) - cos²(x) = 0
thank you steve
To prove the given equation: sin²(3x) + cos²(3x) + sin²(x) - cos²(x) = 0, we can use manipulations of trigonometric identities.
First, let's expand the term sin²(3x) using the triple angle formula:
sin²(3x) = (1 - cos(6x))/2
Next, we can expand the term cos²(3x) using the triple angle formula:
cos²(3x) = (1 + cos(6x))/2
Now, let's rewrite the original equation with the expanded terms:
(1 - cos(6x))/2 + (1 + cos(6x))/2 + sin²(x) - cos²(x) = 0
Combining like terms, we have:
(1 - cos(6x) + 1 + cos(6x))/2 + sin²(x) - cos²(x) = 0
Simplifying further, we get:
(1 + 1)/2 + sin²(x) - cos²(x) = 0
2/2 + sin²(x) - cos²(x) = 0
1 + sin²(x) - cos²(x) = 0
Now, applying the Pythagorean identity sin²(x) + cos²(x) = 1, we can substitute this expression into the equation:
1 + 1 - 1 = 0
So, the left-hand side of the equation simplifies to 1 + 1 - 1 = 0, which confirms that sin²(3x) + cos²(3x) + sin²(x) - cos²(x) is indeed equal to zero.