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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.
Write the expression in terms of sine and cosine and simplify so that no quotients appear in the final expression. cscx(sinx plus cosx)?
csc(x)*{sin(x) + cos(x)} = csc(x)*sin(x) + csc(x)*cos(x) =1/sin*(x)*sin(x) + 1/sin(x)*cos(x) = 1 + cot(x)
Sine cubed plus cosine cubed divided by sine plus cosine?
[sin(x)^3 + cos(x)^3] / [sin(x) + cos(x)]= [(sin(x) + cos(x))(sin(x)^2 - sin(x)cos(x) + cos(x)^2)] / [sin(x) + cos(x)]***Now you can cancel a "sin(x) + cos(x)" from the top and bottom of the fraction. This makes the bottom of the fraction equal to 1. I am just going to write the next step without a 1 on the bottom of the fraction (x/1=x).So now you just have:= (sin(x)^2 - sin(x)cos(x) + cos(x)^2) *I'm going to move some terms around now. ~Not doing any computation in this step.= (sin(x)^2 + cos(x)^2 - sin(x)cos(x)) *Now we know that cos(x)^2 + sin(x)^2 = 1.= 1 - sin(x)cos(x)
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,
Factor sin cubed plus cos cubed?
sin cubed + cos cubed (sin + cos)( sin squared - sin.cos + cos squared) (sin + cos)(1 + sin.cos)
How do you write cosine in terms of sine?
cos(x) = sin(pi/2-x) = -sin(x-pi/2)
Use trigonometric identities to write each expression in terms of a single trigonometric function or a constant. simplify cos t sin t?
Thanks to the pre-existing addition and subtraction theorums, we can establish the identity:sin(a+b) = sin(a)cos(b)+sin(a)cos(b)Then, solving this, we getsin(a+b) = 2(sin(a)cos(b))sin(a)cos(b) = sin(a+b)/2a=b, sosin(a)cos(a) = sin(a+a)/2sin(a)cos(a) = sin(2a)/2Therefore, the answer is sin(2a)/2.
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.
How do you express sin x plus cos x divided by cos x in terms of tan x?
(sin x + cos x) / cosx = sin x / cos x + cosx / cos x = tan x + 1
How do you solve trignometric identities?
tan(x) = sin(x)/cos(x) Therefore, all trigonometric ratios can be expressed in terms of sin and cos. So the identity can be rewritten in terms of sin and cos. Then there are only two "tools": sin^2(x) + cos^2(x) = 1 and sin(x) = cos(pi/2 - x) Suitable use of these will enable you to prove the identity.
How do you use the Euler's formula to obtain the sin3X in terms of cosX and sinX?
Using Euler's Formula, you use (cos(x) + i sin(x))^n = cos (nx) + i sin(nx) Now you let n=3 (cos(x) + i sin (x))3 = cos(3x) + i sin (3x) (cos(x))3 + 3(cos(x))2 * i sin(x) + 3cos(x) * i2 (sin(x))3 = cos(3x)+ i sin(3x) (cos(x))3 + i(3sin(x)(cos (x))2) - 3cos(x)(sin(x)2) - i(sin(x))3 = cos (3x) + i sin(3x) Now only use the terms with i in them to figure out what sin(3x) is... 3sin(x)(cos(x))2 - (sin(x))3 = sin(3x) Hope this helps! :D
Write the expression in terms of sine and cosine and simplify so that no quotients appear in the final expression. cscx(sinx plus cosx)?
csc(x)*{sin(x) + cos(x)} = csc(x)*sin(x) + csc(x)*cos(x) =1/sin*(x)*sin(x) + 1/sin(x)*cos(x) = 1 + cot(x)
How do you apply double angle formulas for csc sec cot?
write in terms of sin, cos or tan then use the double angle formulae. I.e. cosec(x)=1/sin(x) =1/[2sin(x)cos(x)]
Sine cubed plus cosine cubed divided by sine plus cosine?
[sin(x)^3 + cos(x)^3] / [sin(x) + cos(x)]= [(sin(x) + cos(x))(sin(x)^2 - sin(x)cos(x) + cos(x)^2)] / [sin(x) + cos(x)]***Now you can cancel a "sin(x) + cos(x)" from the top and bottom of the fraction. This makes the bottom of the fraction equal to 1. I am just going to write the next step without a 1 on the bottom of the fraction (x/1=x).So now you just have:= (sin(x)^2 - sin(x)cos(x) + cos(x)^2) *I'm going to move some terms around now. ~Not doing any computation in this step.= (sin(x)^2 + cos(x)^2 - sin(x)cos(x)) *Now we know that cos(x)^2 + sin(x)^2 = 1.= 1 - sin(x)cos(x)
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)
Prove that secB - cosB equals tanBsinB?
Try to write everything in terms of sines and cosines:1 / cos B - cos B = (sin B / cos B) sin B1 / cos B - cos B = sin2B / cos BMultiply by the common denominator, cos B:1 - cos2B = sin2BUse the pithagorean identity on the left side:sin2B + cos2B - cos2B = sin2Bsin2B = sin2B
