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In and 8710ABC if sin A and tan A then what is cos A?
In a right triangle, if we know (\sin A) and (\tan A), we can find (\cos A) using the identity (\tan A = \frac{\sin A}{\cos A}). Rearranging this gives us (\cos A = \frac{\sin A}{\tan A}). Therefore, if you have specific values for (\sin A) and (\tan A), you can substitute them into this equation to find (\cos A).
How do you prove this trigonometric relationship sin3A equals 3sinA cos 2 A - sin 3 A?
sin(3A) = sin(2A + A) = sin(2A)*cos(A) + cos(2A)*sin(A)= sin(A+A)*cos(A) + cos(A+A)*sin(A) = 2*sin(A)*cos(A)*cos(A) + {cos^2(A) - sin^2(A)}*sin(A) = 2*sin(A)*cos^2(A) + sin(a)*cos^2(A) - sin^3(A) = 3*sin(A)*cos^2(A) - sin^3(A)
How do I find the product z1z2 if z1 5(cos20 plus isin20) and z2 8(cos15 plus isin15)?
Like normal expansion of brackets, along with: cos(A + B) = cos A cos B - sin A sin B sin(A + B) = sin A cos B + cos A sin B 5(cos 20 + i sin 20) × 8(cos 15 + i sin 15) = 5×8 × (cos 20 + i sin 20)(cos 15 + i sin 15) = 40(cos 20 cos 15 + i sin 15 cos 20 + i cos 15 sin 20 + i² sin 20 sin 15) = 40(cos 20 cos 15 - sin 20 cos 15 + i(sin 15 cos 20 + cos 15 sin 20)) = 40(cos(20 +15) + i sin(15 + 20)) = 40(cos 35 + i sin 35)
How do you simplify cos times cot plus sin?
cos*cot + sin = cos*cos/sin + sin = cos2/sin + sin = (cos2 + sin2)/sin = 1/sin = cosec
How do you simplify cos theta times csc theta divided by tan theta?
'csc' = 1/sin'tan' = sin/cosSo it must follow that(cos) (csc) / (tan) = (cos) (1/sin)/(sin/cos) = (cos) (1/sin) (cos/sin) = (cos/sin)2
In and 8710ABC if sin A and tan A then what is cos A?
In a right triangle, if we know (\sin A) and (\tan A), we can find (\cos A) using the identity (\tan A = \frac{\sin A}{\cos A}). Rearranging this gives us (\cos A = \frac{\sin A}{\tan A}). Therefore, if you have specific values for (\sin A) and (\tan A), you can substitute them into this equation to find (\cos A).
Verify that sin minus cos plus 1 divided by sin plus cos subtract 1 equals sin plus 1 divided by cos?
[sin - cos + 1]/[sin + cos - 1] = [sin + 1]/cosiff [sin - cos + 1]*cos = [sin + 1]*[sin + cos - 1]iff sin*cos - cos^2 + cos = sin^2 + sin*cos - sin + sin + cos - 1iff -cos^2 = sin^2 - 11 = sin^2 + cos^2, which is true,
How do you prove this trigonometric relationship sin3A equals 3sinA cos 2 A - sin 3 A?
sin(3A) = sin(2A + A) = sin(2A)*cos(A) + cos(2A)*sin(A)= sin(A+A)*cos(A) + cos(A+A)*sin(A) = 2*sin(A)*cos(A)*cos(A) + {cos^2(A) - sin^2(A)}*sin(A) = 2*sin(A)*cos^2(A) + sin(a)*cos^2(A) - sin^3(A) = 3*sin(A)*cos^2(A) - sin^3(A)
How would you prove left cosA plus sinA right times left cos2A plus sin2A right equals cosA plus sin3A?
You need to make use of the formulae for sin(A+B) and cos(A+B), and that cos is an even function: sin(A+B) = cos A sin B + sin A cos B cos(A+B) = cos A cos B - sin A sin B cos even fn → cos(-x) = cos(x) To prove: (cos A + sin A)(cos 2A + sin 2A) = cos A + sin 3A The steps are to work with the left hand side, expand the brackets, collect [useful] terms together, apply A+B formula above (backwards) and apply even nature of cos function: (cos A + sin A)(cos 2A + sin 2A) = cos A cos 2A + cos A sin 2A + sin A cos 2A + sin A sin 2A = (cos A cos 2A + sin A sin 2A) + (cos A sin 2A + sin A cos 2A) = cos(A - 2A) + sin(A + 2A) = cos(-A) + sin 3A = cos A + sin 3A which is the right hand side as required.
Prove that sin 90 equals cos 50sin 40- cos 40 sin 50?
Sorry, but cos(50)sin(40) - cos(40)sin(50) is -0.1736, which is not even close to sin(90) which is 1.This does not work in radians, either. Please restate your question.
Which trig expression is equal to sin (72 and Acirc and deg - a)?
The expression ( \sin(72^\circ - a) ) can be rewritten using the sine difference identity: [ \sin(72^\circ - a) = \sin(72^\circ) \cos(a) - \cos(72^\circ) \sin(a). ] Thus, ( \sin(72^\circ - a) ) is equal to ( \sin(72^\circ) \cos(a) - \cos(72^\circ) \sin(a) ).
How do you show that 2 sin squared x minus 1 divided by sin x minus cos x equals sin x plus cos x?
(2 sin^2 x - 1)/(sin x - cos x) = sin x + cos x (sin^2 x + sin^2 x - 1)/(sin x - cos x) =? sin x + cos x [sin^2 x - (1 - sin^2 x)]/(sin x - cos x) =? sin x + cos x (sin^2 x - cos^2 x)/(sin x - cos x) =? sin x + cos x [(sin x - cos x)(sin x + cos x)]/(sin x - cos x) =? sin x + cos x sin x + cos x = sin x + cos x
How do I find the product z1z2 if z1 5(cos20 plus isin20) and z2 8(cos15 plus isin15)?
Like normal expansion of brackets, along with: cos(A + B) = cos A cos B - sin A sin B sin(A + B) = sin A cos B + cos A sin B 5(cos 20 + i sin 20) × 8(cos 15 + i sin 15) = 5×8 × (cos 20 + i sin 20)(cos 15 + i sin 15) = 40(cos 20 cos 15 + i sin 15 cos 20 + i cos 15 sin 20 + i² sin 20 sin 15) = 40(cos 20 cos 15 - sin 20 cos 15 + i(sin 15 cos 20 + cos 15 sin 20)) = 40(cos(20 +15) + i sin(15 + 20)) = 40(cos 35 + i sin 35)
How do you prove that 2 sin 3x divided by sin x plus 2 cos 3x divided by cos x equals 8 cos 2x?
You need to know the trigonometric formulae for sin and cos of compound angles. sin(x+y) = sin(x)*cos(y)+cos(x)*sin(y) and cos(x+y) = cos(x)*cos(y) - sin(x)*sin(y) Using these, y = x implies that sin(2x) = sin(x+x) = 2*sin(x)cos(x) and cos(2x) = cos(x+x) = cos^2(x) - sin^2(x) Next, the triple angle formulae are: sin(3x) = sin(2x + x) = 3*sin(x) - 4*sin^3(x) and cos(3x) = 4*cos^3(x) - 3*cos(x) Then the left hand side = 2*[3*sin(x) - 4*sin^3(x)]/sin(x) + 2*[4*cos^3(x) - 3*cos(x)]/cos(x) = 6 - 8*sin^2(x) + 8cos^2(x) - 6 = 8*[cos^2(x) - sin^2(x)] = 8*cos(2x) = right hand side.
What is the integral of sin3ycos5ydy?
Best way: Use angle addition. Sin(Ax)Cos(Bx) = (1/2) [sin[sum x] + sin[dif x]], where sum = A+B and dif = A-B To show this, Sin(Ax)Cos(Bx) = (1/2) [sin[(A+B) x] + sin[(A-B) x]] = (1/2) [(sin[Ax]Cos[Bx]+sin[Bx]cos[Ax]) + (sin[Ax]cos[-Bx]+sin[-Bx]cos[Ax])] Using the facts that cos[-k] = cos[k] and sin[-k] = -sin[k], we have: (1/2) [(sin[Ax]Cos[Bx]+sin[Bx]cos[Ax]) + (sin[Ax]cos[-Bx]+sin[-Bx]cos[Ax])] (1/2) [(sin[Ax]Cos[Bx]+sin[Bx]cos[Ax]) + (sin[Ax]cos[Bx]-sin[Bx]cos[Ax])] (1/2) 2sin[Ax]Cos[Bx] sin[Ax]Cos[Bx] So, Int[Sin(3y)Cos(5y)dy] = (1/2)Int[Sin(8y)-Sin(2y)dy] = (-1/16) Cos[8y] +1/4 Cos[2y] + C You would get the same result if you used integration by parts twice and played around with trig identities.
How do you simplify cos times cot plus sin?
cos*cot + sin = cos*cos/sin + sin = cos2/sin + sin = (cos2 + sin2)/sin = 1/sin = cosec
Factor sin cubed plus cos cubed?
sin cubed + cos cubed (sin + cos)( sin squared - sin.cos + cos squared) (sin + cos)(1 + sin.cos)
