prove the ff:
a.) sin(x/2) - cos(x/2) = +-sqrt(1 - sin(x))
b.) tan(x/2) + cot(x/2) = 2csc(x)
let x/2 = Ø , then your equation becomes
sinØ - cosØ = ±√(1 - sin 2Ø)
we know that
if LS = RS , then
(LS)^2 = (RS)^2
LS = sinØ - cosØ
(LS)^2 = (sinØ - cosØ)^2
= sin^2 Ø - 2sinØcosØ + cos^2 Ø
= 1 - sin 2Ø
= (RS)^2 , could stop here, but carry on ....
LS = ± √(1 - sin 2Ø)
= RS
2nd part
I will make a similar substitution
let x/2 = y, then we have
prove: tany + coty = 2cscy
LS = siny/cosy + cosy/siny
= (sin^2 y + cos^2 y)/sinycosy
but (2sinycosy = siny)
= 1/((1/2)siny)
= 2cscy
= 2csc(x/2)
= RS
To prove these trigonometric identities, we will use the fundamental trigonometric identities and algebraic manipulation. Let's start with proving each of them separately:
a.) Proof of sin(x/2) - cos(x/2) = ±√(1 - sin(x)):
Step 1: Start with the left side of the equation and apply the double-angle identities for sine and cosine:
sin(x/2) - cos(x/2)
= 2sin(x/2)cos(x/2) - 2cos^2(x/2) + 1
Step 2: Simplify and group terms together:
= 2sin(x/2)cos(x/2) - (2cos^2(x/2) - 1)
= 2sin(x/2)cos(x/2) - (1 - sin^2(x/2))
= 2sin(x/2)cos(x/2) - sin^2(x/2)
Step 3: Factor out sin(x/2) from the first term:
= sin(x/2)(2cos(x/2) - sin(x/2))
Step 4: Apply the Pythagorean identity for sine and simplify:
= sin(x/2)(√(1 - sin^2(x/2)) - sin(x/2))
= sin(x/2)√(1 - sin^2(x/2)) - sin^2(x/2)
Step 5: Simplify the second term by applying the Pythagorean identity for sine squared:
= sin(x/2)√(1 - sin^2(x/2)) - (1 - cos^2(x/2))
Step 6: Apply the Pythagorean identity for cosine squared:
= sin(x/2)√(1 - sin^2(x/2)) - (1 - (1 - sin^2(x/2)))
Step 7: Simplify and combine like terms:
= sin(x/2)√(1 - sin^2(x/2)) - sin^2(x/2) + sin^2(x/2)
= sin(x/2)√(1 - sin^2(x/2))
Step 8: Finally, since √(1 - sin^2(x/2)) is the same as cos(x/2), we have:
= sin(x/2)cos(x/2)
= √(1 - sin^2(x/2))
Therefore, sin(x/2) - cos(x/2) = ±√(1 - sin(x)), where the ± symbol accounts for the positive and negative values of the square root.
b.) Proof of tan(x/2) + cot(x/2) = 2csc(x):
Step 1: Start with the left side of the equation and express tangent and cotangent in terms of sine and cosine:
tan(x/2) + cot(x/2)
= (sin(x/2)/cos(x/2)) + (cos(x/2)/sin(x/2))
Step 2: Find a common denominator for the two fractions:
= (sin(x/2)sin(x/2))/(cos(x/2)sin(x/2)) + (cos(x/2)cos(x/2))/(sin(x/2)cos(x/2))
Step 3: Simplify the fractions:
= sin^2(x/2)/(cos(x/2)sin(x/2)) + cos^2(x/2)/(sin(x/2)cos(x/2))
Step 4: Multiply each fraction by the reciprocal of the denominator:
= sin^2(x/2)/(cos(x/2)sin(x/2)) + cos^2(x/2)/(sin(x/2)cos(x/2))
= sin^2(x/2)/(sin(x/2)cos(x/2)) + cos^2(x/2)/(sin(x/2)cos(x/2))
= (sin^2(x/2) + cos^2(x/2))/(sin(x/2)cos(x/2))
Step 5: Apply the Pythagorean identity for sine and cosine:
= 1/(sin(x/2)cos(x/2))
Step 6: Apply the reciprocal identity for sine and cosine:
= 1/sin(x/2) * 1/cos(x/2)
= csc(x/2) * sec(x/2)
Step 7: Apply the double-angle identity for cosecant and secant:
= csc(x)sec(x)
= 2csc(x)
Therefore, tan(x/2) + cot(x/2) = 2csc(x).
Both identities have now been proven.