Prove:
sin2x / 1 - cos2x = cotx
My Attempt:
LS:
= 2sinxcosx / - 1 - (1 - 2sin^2x)
= 2sinxcosx / - 1 + 2sin^2x
= cosx / sinx - 1
= cosx / sinx - 1/1
= cosx / sinx - sinx / sinx
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Prove:
2sin(x+y)sin(x-y) = cos2y - cos2x
My Attempt:
RS:
= 1 - 2sin^2y - 1 - 2sin^2x
= 1 - 1 - 2sin^2y - 2sin^2x
= -2sin^2y - 2sin^2x
Solved the first problem, I know what I did wrong...
LS:
= 2sinxcosx /1 - (1 - 2sin^2x)
= 2sinxcosx / 1 - 1 + 2sin^2x
= 2sinxcosx / 2sin^2x
= cosx / sinx
= cotx
sin2x / 1 - cos2x = cotx
2sinxcosx / 1 - (1-2sin^2x)
2sinxcosx / 2sin^2x
cosx/sinx = cotx
I know how to solve the first question.
2sin(x+y)sin(x-y) = cos2y - cos2x
lhs
2(sinx cosy + cosx siny) (sin x cos y – cosx siny)
2( sin^2xcos^2y – sinxcosycosxsiny + sinxcosycosxsiny – cos^2xsin^2y)
2(sin^2xcos^2y – cos^2xsin^2y)
2[(1-cos^2x)cos^2y – (1-cos^2y)cos^2x]
2[cos^2y-cos^2xcos^2y – cos^2x + cos^2xcos^2y]
2[cos^2y-cos^2x]
rhs
2 cos^2y - 1 - 2cos^2x+1
2[cos^2y – cos^2x]
thanks
To prove the given trigonometric identity, we can start by manipulating the left-hand side (LHS) and right-hand side (RHS) separately until they are equal.
Proving the identity: sin(2x) / (1 - cos(2x)) = cot(x)
First, let's simplify the left-hand side (LHS):
LHS = sin(2x) / (1 - cos(2x))
Using the double angle formula for sine (sin(2x) = 2sin(x)cos(x)) and the double angle formula for cosine (cos(2x) = cos^2(x) - sin^2(x)), we can rewrite the LHS as:
LHS = 2sin(x)cos(x) / (1 - (cos^2(x) - sin^2(x)))
LHS = 2sin(x)cos(x) / (1 - cos^2(x) + sin^2(x))
LHS = 2sin(x)cos(x) / (sin^2(x) + (1 - cos^2(x)))
Next, we can simplify the denominator:
LHS = 2sin(x)cos(x) / (sin^2(x) + sin^2(x))
LHS = 2sin(x)cos(x) / (2sin^2(x))
LHS = cos(x) / sin(x)
Using the reciprocal identity (cot(x) = cos(x) / sin(x)), we see that LHS is equal to cot(x), which proves the identity.
Therefore, sin(2x) / (1 - cos(2x)) = cot(x).
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Proving the identity: 2sin(x+y)sin(x-y) = cos(2y) - cos(2x)
Starting with the right-hand side (RHS):
RHS = cos(2y) - cos(2x)
Using the double angle formula for cosine, we have:
RHS = cos^2(y) - sin^2(y) - (cos^2(x) - sin^2(x))
RHS = sin^2(x) - sin^2(y) - cos^2(x) + cos^2(y)
Now, let's focus on the left-hand side (LHS):
LHS = 2sin(x+y)sin(x-y)
Using the product-to-sum trigonometric identity (sin(A)sin(B) = 1/2 [cos(A-B) - cos(A+B)]), we can rewrite the LHS as:
LHS = 2 * (1/2)[cos((x+y)-(x-y)) - cos((x+y)+(x-y))]
LHS = cos(2y) - cos(2x)
LHS and RHS are equal, so the identity is proved:
2sin(x+y)sin(x-y) = cos(2y) - cos(2x)