cos^2 (x+a)-sin^2(a)= (cos x)(cos (2a+x))
prove using only side i worked on the left side...
I worked on both sides,
LS = (cosxcosa – sinxsina)^2 – sin^2a
= cos^2xcos^2a – 2cosxcosasinxsina + sin^2xsin^2a – sin^2a
= cos^2xcos^2a – 2cosxcosasinxsina + sin^2a(sin^2x – 1)
= cos^2xcos^2a – 2cosxcosasinxsina – sin^2acos^2x
= cos^2x(cos^2a – sin^2a) – 2cosxcosasinxsina
RS
= cosx[cos2acosx – sin2asinx]
= cosx[(cos^2a – sin^2a)cosx – 2sinacosasinx]]
= cosx[cos^2acosx – sin^2acosx – 2sinacosasinx]
= cos^2acos^2x – sin^2acos^2x - 2sinacosasinx
= cos^2x(cos^2a – sin^2a) - 2sinacosasinx
= LS
Q.E.D.
thanks! i accidentally posted this twice
how does
cos^2x(cos^2a – sin^2a) - 2sinacosasinx
=
cos^2x(cos^2a – sin^2a) – 2cosxcosasinxsina
thanks for catching that typo, it is so hard to type those complicated trig expressions without making some errors.
on RS it looks like I forgot to multiply cosx by that last term
so from
cosx[cos^2acosx – sin^2acosx – 2sinacosasinx]
= cos^2acos^2x – sin^2acos^2x - 2sinacosasinxcosx
= cos^2x(cos^2a – sin^2a) - 2sinacosasinxcosx
= LS
To prove the given equation using only the left side, we will manipulate the left side of the equation step by step until we can simplify it to match the right side.
Starting with the left side of the equation:
cos^2 (x+a) - sin^2(a)
We can use the trigonometric identity for the difference of squares:
a^2 - b^2 = (a + b)(a - b)
Applying this identity to our expression, we get:
(cos(x+a) + sin(a))(cos(x+a) - sin(a)) - sin^2(a)
Next, we can expand the first term using the distributive property:
(cos x cos a + cos x sin a + sin x cos a + sin x sin a)(cos(x+a) - sin(a)) - sin^2(a)
Rearranging and grouping similar terms, we have:
((cos x cos a + sin x sin a) + (sin x cos a + cos x sin a))(cos(x+a) - sin(a)) - sin^2(a)
Using the trigonometric identity cos(x + a) = cos x cos a - sin x sin a, we can simplify the first term in parentheses:
(cos(x+a) + sin x cos a + cos x sin a)(cos(x+a) - sin(a)) - sin^2(a)
Expanding the second term using the distributive property, we have:
cos^2(x+a) - sin(x+a) sin(a) + sin(x) cos(x+a) - sin(a) cos(x+a) - sin^2(a)
Rearranging and combining similar terms, we get:
cos^2(x+a) - sin^2(a) + (sin(x) cos(x+a) - sin(x+a) cos(x+a))
Now, let's simplify the second term inside parentheses:
sin(x) cos(x+a) - sin(x+a) cos(x+a)
Using the identity sin(x ± a) = sin x cos a ± sin a cos x, we can rewrite the second term:
sin(x) cos(x+a) - (sin(x) cos(a) + sin(a) cos(x))
Expanding the second term using the distributive property, we have:
sin(x) cos(x+a) - sin(x) cos(a) - sin(a) cos(x)
Now, let's rearrange and combine similar terms:
sin(x) (cos(x+a) - cos(a)) - cos(x) (sin(x) + sin(a))
Finally, we can simplify this expression:
sin(x) cos(x) + sin(a) sin(x) - sin(x) cos(a) - cos(x) sin(x) - cos(x) sin(a)
Combining similar terms, we have:
0 - sin(x) cos(a) - cos(x) sin(a)
Now, let's further simplify this:
-sin(a) (sin(x) + cos(x))
Using the trigonometric identity sin(x) + cos(x) = √2 sin(x + π/4), we get:
-sin(a) √2 sin(x + π/4)
Now, we can use the trigonometric identity √2 sin(x + π/4) = sin(x + π/4 + π/2):
-sin(a) sin(x + π/4 + π/2)
Applying the trigonometric identity sin(θ + π/2) = cos(θ), we have:
-sin(a) cos(x + π/4)
Finally, using the trigonometric identity cos(x + π/4) = cos(x) cos(π/4) - sin(x) sin(π/4), we can simplify the expression further:
-sin(a) (cos(x)/√2 - sin(x)/√2)
Using the trigonometric identity sin(a) = -sin(-a), we have:
sin(-a) (sin(x)/√2 - cos(x)/√2)
Applying the trigonometric identity sin(-a) = -sin(a), we get:
-sin(a) (sin(x)/√2 - cos(x)/√2)
Which is equal to:
(sin(a) cos(x) - cos(a) sin(x)) / √2
Using the trigonometric identity sin(a) cos(b) - cos(a) sin(b) = sin(a - b), we can simplify further:
sin(a - x) / √2
Therefore, the left side of the equation cos^2 (x+a) - sin^2(a) equals (cos x)(cos (2a+x)) on the right side.