Hence proved the statement that is [tex]\frac{sin4\theta+sin3 \theta}{1+cos2\theta+cos4\theta}= tan 2\theta[/tex] with formulas of trigonometry sin A+sin B = 2 sin ([tex]\frac{A+B}{2}[/tex]) cos([tex]\frac{A-B}{2}[/tex]) and cos2θ = 2cos²θ-1.
Given that,
We have to prove that [tex]\frac{sin4\theta+sin3 \theta}{1+cos2\theta+cos4\theta}= tan 2\theta[/tex]
We know that,
What is trigonometric ratio?
The sine, cosine, tangent, secant, cosecant, and cotangent are six trigonometric ratios used in trigonometry. These ratios are abbreviated as sin, cos, tan, sec, cosec, and cot, respectively. Let's examine the right-angled triangle displayed below. Any two sides of a right-angled triangle, which have a total of three sides, can be compared in terms of their relative angles using trigonometric ratios.
[tex]\frac{sin4\theta+sin3 \theta}{1+cos2\theta+cos4\theta}= tan 2\theta[/tex]
We know take the LHS
sinA+sinB = 2 sin ([tex]\frac{A+B}{2}[/tex]) cos([tex]\frac{A-B}{2}[/tex]) and cos2θ = 2cos²θ-1
Therefore, hence proved the statement that is [tex]\frac{sin4\theta+sin3 \theta}{1+cos2\theta+cos4\theta}= tan 2\theta[/tex] with formulas of trigonometry sin A + sin B = 2 sin ([tex]\frac{A+B}{2}[/tex]) cos([tex]\frac{A-B}{2}[/tex]) and cos2θ = 2cos²θ-1.
Answers & Comments
Step-by-step explanation:
(Sin4θ + Sin2θ)/(1 + Cos2θ +Cos4θ) = Tan2θ
LHS = (Sin4θ + Sin2θ)/(1 + Cos2θ +Cos4θ)
Using Sin2x = 2SinxCosx => Sin4θ = 2Sin2θCos2θ
& Cos2x = 2Cos²x - 1 => Cos4θ = 2Cos²2θ - 1
= (2Sin2θCos2θ + Sin2θ)/(1 + Cos2θ + 2Cos²2θ - 1)
= Sin2θ(2Cos2θ + 1) /(Cos2θ + 2Cos²2θ)
= Sin2θ(2Cos2θ + 1) /Cos2θ(1 + 2Cos2θ)
= Sin2θ(1 + 2Cos2θ) /Cos2θ(1 + 2Cos2θ)
Cancelling 1 + 2Cos2θ
= Sin2θ /Cos2θ
= Tan2θ
= RHS
Hence proved the statement that is [tex]\frac{sin4\theta+sin3 \theta}{1+cos2\theta+cos4\theta}= tan 2\theta[/tex] with formulas of trigonometry sin A+sin B = 2 sin ([tex]\frac{A+B}{2}[/tex]) cos([tex]\frac{A-B}{2}[/tex]) and cos2θ = 2cos²θ-1.
Given that,
We have to prove that [tex]\frac{sin4\theta+sin3 \theta}{1+cos2\theta+cos4\theta}= tan 2\theta[/tex]
We know that,
What is trigonometric ratio?
The sine, cosine, tangent, secant, cosecant, and cotangent are six trigonometric ratios used in trigonometry. These ratios are abbreviated as sin, cos, tan, sec, cosec, and cot, respectively. Let's examine the right-angled triangle displayed below. Any two sides of a right-angled triangle, which have a total of three sides, can be compared in terms of their relative angles using trigonometric ratios.
[tex]\frac{sin4\theta+sin3 \theta}{1+cos2\theta+cos4\theta}= tan 2\theta[/tex]
We know take the LHS
sinA+sinB = 2 sin ([tex]\frac{A+B}{2}[/tex]) cos([tex]\frac{A-B}{2}[/tex]) and cos2θ = 2cos²θ-1
[tex]\frac{2sin3\theta\times cos\theta}{1+cos2\theta+2cos^{2}2\theta-1 }[/tex]
[tex]\frac{2cos\theta(sin3\theta)}{cos2\theta(1+2cos2\theta) }[/tex]
sin3θ = 3sinθ-4sin³θ
[tex]\frac{2cos\theta(3sin\theta-4sin^{3}\theta )}{cos2\theta(1+2cos2\theta) }[/tex]
[tex]\frac{2cos\theta sin\theta(3-4sin^{2}\theta )}{cos2\theta(1+2-4sin^{2}\theta) }[/tex]
[tex]\frac{2cos\theta sin\theta(3-4sin^{2}\theta )}{cos2\theta(3-4sin^{2}\theta) }[/tex]
[tex]\frac{2cos\theta sin\theta}{cos2\theta}[/tex]
tan2θ
Therefore, hence proved the statement that is [tex]\frac{sin4\theta+sin3 \theta}{1+cos2\theta+cos4\theta}= tan 2\theta[/tex] with formulas of trigonometry sin A + sin B = 2 sin ([tex]\frac{A+B}{2}[/tex]) cos([tex]\frac{A-B}{2}[/tex]) and cos2θ = 2cos²θ-1.
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