sin4A = sin(2A + 2A) We know that, sin(A + B) = sinA cosB + cosA sinB. Therefore, sin4A = sin2A cos2A + cos2A sin2A. ⇒ sin4A = 2sin2A cos2A. From T-ratios of multiple angles, We get, sin2A = 2sinA cosA and cos2A = cos 2 A – sin 2 A. ⇒ sin4A = 2(2sinA cosA)(cos 2 A – sin 2 A) ⇒ sin4A = 4sinA cos 3 A – 4cosA sin 3 A. Hence, sin4A ...

Apply the double angle formula twice for sin, and you get an expression sin(4a) = 4cos(a)sin(a)cos(2a) the first factor of which is zero - you don't need to expand the cos(2a) bit. Share Cite

prove\:\tan^2(x)-\sin^2(x)=\tan^2(x)\sin^2(x) \frac{d}{dx}(\frac{3x+9}{2-x}) (\sin^2(\theta))' \sin(120) \lim _{x\to 0}(x\ln (x)) \int e^x\cos (x)dx \int_{0}^{\pi}\sin(x)dx

What are sin(4a) and cos(4a) if tan(a) = 3? The desired trig function values are sin (4a) = -24/25; cos (4a) = 7/25. The values are sin(4a) = −2524 and cos(4a) = 257. To find these, we used the known value of tan(a) = 3 to calculate sin(a) and cos(a), followed by …

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Sep 2, 2020 · sin 4A = sin (2A + 2A) We know that, sin(A + B) = sin A cos B + cos A sin B. Therefore, sin 4A = sin 2A cos 2A + cos 2A sin 2A. ⇒ sin 4A = 2 sin 2A cos 2A. From T-ratios of multiple angle, We get, sin 2A = 2 sin A cos A and cos 2A = cos 2 A – sin 2 A. ⇒ sin 4A = 2(2 sin A cos A)(cos 2 A – sin 2 A) ⇒ sin 4A = 4 sin A cos 3 A – 4 cos ...

Jul 7, 2024 · \[ \sin 4A = 4 \sin A \cos^3 A - 4 \cos A \sin^3 A \] we can start by using the angle addition formula for sine and then apply trigonometric identities. Let's break it down step-by-step. First, we use the double-angle formulas for sine: \[ \sin 4A = \sin (2 \cdot 2A) = 2 \sin 2A \cos 2A \]

Dec 29, 2017 · sin 4A =sin2(2A); =2sin2Acos2A; since, sin2A=2sinAcosA: =2(2sinAcosA)cos2A; =4sinAcosAcos2A; since, cos2A=cos²A-sin²A: =4sinAcosA(cos²A-sin²A); =4sinAcos³A-4sin³AcosA; =4sinAcos³A-4cosAsin³A bye :-)

Use the Double Angle identity: sin (2 x) = 2 sin (x) cos (x) = 2 ⋅ 2 sin (a) cos (a) cos (2 a)

From the right triangle in the diagram with sin A = 4/5, Pythagoras' Theorem shows that the third side of the triangle has length 3 and hence cos A = 3/5. Use the exprression . sin 2A = 2 sin A cos A to calculate sin 2A. Now start again. Since you know sin 2A use Pythagoras' Theorem to find cos 2A and hence calculate sin 4A. Cheers Harley