EXERCISE 8 1. Prove that the following statements are identities. $$ \begin{array}{ll} ext { (a) }(\sin heta+\cos heta)^{2}=1+2 \sin heta \cdot \cos heta & ext { (An important result to remember) } \ ext { (b) } \frac{1}{ an heta}=\frac{\cos heta}{\sin heta} & ext { (An important result to remember) } \ ext { (c) } \frac{1}{\sin ^{2} heta}-\frac{\cos ^{2} heta}{\sin ^{2} heta}=1 & ext { (d) } \frac{ an x \cdot \cos x}{\sin x}=1 \ ext { (e) } \sin ^{4} x-\cos ^{4} x=\sin ^{2} x-\cos ^{2} x & ext { (f) } \sin ^{2} heta+\sin ^{2} heta \cdot an ^{2} heta= an ^{2} heta \ ext { (g) } \cos heta(1+ an heta)=\cos heta+\sin heta & ext { (h) } \frac{1-\cos ^{2} heta}{\cos ^{2} heta+2 \cos heta+1}=\frac{1-\cos heta}{1+\cos heta} \ ext { (i) } \frac{1}{1-\cos heta}+\frac{1}{1+\cos heta}=\frac{2}{\sin ^{2} heta} & ext { (j) } \frac{1}{\sin heta}+\frac{1}{ an heta}=\frac{1+\cos heta}{\sin \left(180^{\circ}- heta ight)}\end{array} $$

Question

EXERCISE 8 1. Prove that the following statements are identities. $$ \begin{array}{ll}	ext { (a) }(\sin 	heta+\cos 	heta)^{2}=1+2 \sin 	heta \cdot \cos 	heta & 	ext { (An important result to remember) } \ 	ext { (b) } \frac{1}{	an 	heta}=\frac{\cos 	heta}{\sin 	heta} & 	ext { (An important result to remember) } \ 	ext { (c) } \frac{1}{\sin ^{2} 	heta}-\frac{\cos ^{2} 	heta}{\sin ^{2} 	heta}=1 & 	ext { (d) } \frac{	an x \cdot \cos x}{\sin x}=1 \ 	ext { (e) } \sin ^{4} x-\cos ^{4} x=\sin ^{2} x-\cos ^{2} x & 	ext { (f) } \sin ^{2} 	heta+\sin ^{2} 	heta \cdot 	an ^{2} 	heta=	an ^{2} 	heta \ 	ext { (g) } \cos 	heta(1+	an 	heta)=\cos 	heta+\sin 	heta & 	ext { (h) } \frac{1-\cos ^{2} 	heta}{\cos ^{2} 	heta+2 \cos 	heta+1}=\frac{1-\cos 	heta}{1+\cos 	heta} \ 	ext { (i) } \frac{1}{1-\cos 	heta}+\frac{1}{1+\cos 	heta}=\frac{2}{\sin ^{2} 	heta} & 	ext { (j) } \frac{1}{\sin 	heta}+\frac{1}{	an 	heta}=\frac{1+\cos 	heta}{\sin \left(180^{\circ}-	hetaight)}\end{array} $$

UpStudy AI · Accepted Answer

Below is a step‐by‐step explanation for each of the parts (a)–(j).

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(a) Prove (sin θ + cos θ)² = 1 + 2 sin θ · cos θ

• Start by expanding the left-hand side (LHS):
  (sin θ + cos θ)² = sin²θ + 2 sin θ cos θ + cos²θ

• Recall the Pythagorean identity:
  sin²θ + cos²θ = 1

• Thus, the expansion becomes:
  1 + 2 sin θ cos θ

• This is exactly the right-hand side (RHS).

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(b) Prove 1/ tan θ = cos θ/ sin θ

• By definition, tan θ = sin θ/ cos θ (assuming cos θ ≠ 0).

• Then,
  1/ tan θ = 1/(sin θ/ cos θ) = cos θ/ sin θ

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(c) Prove 1/sin²θ – cos²θ/sin²θ = 1

• Combine the two terms over a common denominator:
  1/sin²θ – cos²θ/sin²θ = (1 – cos²θ)/sin²θ

• Again, using the Pythagorean identity:
  1 – cos²θ = sin²θ

• Thus, we have:
  sin²θ/sin²θ = 1

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(d) Prove (tan x · cos x)/ sin x = 1

• Recall that tan x = sin x/ cos x (provided cos x ≠ 0).

• Substitute into the expression:
  (tan x · cos x)/ sin x = [(sin x/ cos x) · cos x]/ sin x = sin x/sin x = 1

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(e) Prove sin⁴ x – cos⁴ x = sin² x – cos² x

• Recognize the left-hand side as a difference of two squares:
  sin⁴ x – cos⁴ x = (sin² x)² – (cos² x)² 
    = (sin² x – cos² x)(sin² x + cos² x)

• Using sin² x + cos² x = 1, this reduces to:
  (sin² x – cos² x)·1 = sin² x – cos² x

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(f) Prove sin² θ + sin² θ · tan² θ = tan² θ

• Factor sin² θ from the left-hand side:
  sin² θ + sin² θ · tan² θ = sin²θ (1 + tan²θ)

• Recall the trigonometric identity:
  1 + tan²θ = sec²θ = 1/ cos²θ

• Thus, the expression becomes:
  sin²θ (1/ cos²θ) = sin²θ/ cos²θ = tan²θ

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(g) Prove cos θ (1 + tan θ) = cos θ + sin θ

• Expand the left-hand side:
  cos θ (1 + tan θ) = cos θ + cos θ · tan θ

• Substitute tan θ = sin θ/ cos θ:
  cos θ + cos θ · (sin θ/ cos θ) = cos θ + sin θ

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(h) Prove (1 – cos²θ) / (cos²θ + 2 cos θ + 1) = (1 – cos θ)/(1 + cos θ)

• Notice that the numerator can be factored using the Pythagorean identity:
  1 – cos²θ = (1 – cos θ)(1 + cos θ)

• Also, the denominator factors as a perfect square:
  cos²θ + 2 cos θ + 1 = (cos θ + 1)²

• Thus, the expression becomes:
  [(1 – cos θ)(1 + cos θ)]/(1 + cos θ)² = (1 – cos θ)/(1 + cos θ)
  (provided cos θ ≠ –1 so that 1 + cos θ ≠ 0)

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(i) Prove 1/(1 – cos θ) + 1/(1 + cos θ) = 2/ sin²θ

• Add the two fractions by finding a common denominator:
  (1/(1 – cos θ)) + (1/(1 + cos θ)) =
  [ (1 + cos θ) + (1 – cos θ) ] / [ (1 – cos θ)(1 + cos θ) ]

• Simplify the numerator:
  (1 + cos θ) + (1 – cos θ) = 2

• Recognize the denominator as a difference of squares:
  (1 – cos θ)(1 + cos θ) = 1 – cos²θ = sin²θ

• Thus, the expression becomes:
  2/ sin²θ

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(j) Prove 1/sin θ + 1/tan θ = (1 + cos θ)/ sin(180° – θ)

• Start with the left-hand side. Recall from part (b) that:
  1/tan θ = cos θ/ sin θ

• So,
  1/sin θ + 1/tan θ = 1/sin θ + cos θ/ sin θ = (1 + cos θ)/ sin θ

• Recall that sin(180° – θ) = sin θ for any angle θ, so:
  (1 + cos θ)/ sin θ = (1 + cos θ)/ sin(180° – θ)

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Thus, each of the given expressions has been shown to be identically equal to its counterpart.
All the given trigonometric identities have been proven to be true.

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