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d/dx (sin x)^2=2sinxcosx
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1-2sin squared x equals?
3
How does 1-sin squared x times 1 plus tan suared x equals 1?
Use these identities: sin2(x) + cos2(x) = 1, and tan(x) = sin(x)/cos(x) For clarity, the functions are written here without their arguments (the "of x" part). (1 - sin2) = cos2 (1 + tan2) = (1 + sin2/cos2) = (cos2+sin2) / cos2 = 1/cos2 Multiply them: (cos2) times (1/cos2) = 1'QED'
How do you figure out the Trigonometry theorem?
If, by trigonometry theorem you mean the "fundamental theorem of trigonometry," sin2(x) + cos2(x) = 1, it is actually the Pythagorean Theorem. if you have a right triangle with a hypotenuse of one, sin(x) is one leg, and cos(x) is the other. The Pythagorean Theorem states that a2 + b2 = c2 and therefore sin2(x) + cos2(x) = 1.
Express sin 2x as a function of sin x?
The double angle formula states that sin(2x) = 2sin(x)cos(x), and because sin2(x) + cos2(x) = 1, sin(2x) = 2sin(x)*SQRT(1-sin2(x)). However, that will only give you the correct result when cos(x) is positive. (because SQRT only returns the positive square root)
What is sine theta plus 2 cosine squared theta?
I shall use x instead of theta since I cannot be bothered to paste it at each step.sin(x) + 2*cos2(x) = sin(x) + 2*[1 - sin2(x)] = sin(x) + 2 - 2sin2(x) which cannot be simplified further.I shall use x instead of theta since I cannot be bothered to paste it at each step.sin(x) + 2*cos2(x) = sin(x) + 2*[1 - sin2(x)] = sin(x) + 2 - 2sin2(x) which cannot be simplified further.I shall use x instead of theta since I cannot be bothered to paste it at each step.sin(x) + 2*cos2(x) = sin(x) + 2*[1 - sin2(x)] = sin(x) + 2 - 2sin2(x) which cannot be simplified further.I shall use x instead of theta since I cannot be bothered to paste it at each step.sin(x) + 2*cos2(x) = sin(x) + 2*[1 - sin2(x)] = sin(x) + 2 - 2sin2(x) which cannot be simplified further.
What is the derivative of cot x?
The derivative of cot(x) is -csc2(x).(Which is the same as -1/sin2(x).)
Find the second derivative of y equals sin sqaured x?
y = sin2(x) y' = 2sin(x)cos(x) y'' = 2 [ cos(x)cos(x) + sin(x)(-sin(x)) ] = 2 [ cos2(x) - sin2(x) ] = 2 [ 1 - sin2(x) - sin2(x) ] = 2 [ 1 - 2sin2(x) ]
Cos x - cos x sin2 x equals cos3x?
cos(x)-cos(x)sin2(x)=[cos(x)][1-sin2(x)]cos(x)-cos(x)sin2(x)=[cos(x)][cos2(x)]cos(x)-cos(x)sin2(x)=cos3(x)
What is sin squared minus 1?
-cos2(x)1 = sin2(x) +cos2(x)1 - cos2(x) = sin2(x)-cos2(x) = sin2(x) - 1
1-2sin squared x equals?
3
Can you show 1 divided by 1 plus cosx equals cscsquaredx minus cscxcotx?
2
What is the derivative of 1 divided by sinx cosx?
use the chain rule and product rule: f(x)=1/u(x), f'(x)=-1/u2(x)*u'(x) and f(x)=a(x)b(x), f'(x)=a'(x)b(x)+b'(x)a(x) the derivative of sin is cos, and the derivative of cos in negative sin. so f(x)=1/(sin(x)cos(x)), f'(x)=-1/(sin(x)cos(x))2*(cos2(x)-sin2(x)) or assume sin(x)cos(x)=.5sin(2x) so using chain rule, f'(x)=-2/sin2(2x)*.5cos(2x)*2. Both answers are equivalent.
What is 2 cos squared x - sinx - 1?
1
How can you write an expression for cos x in term of sin x using pythagorean theorem?
cos2 x + sin2 x = 1 cos2 x = 1 - sin2 x
Sec²x - tan²x equals?
sec2(x) - tan2(x)= 1/cos2(x) - sin2(x)/cos2(x)= (1 - sin2(x)) / cos2(x)= cos2(x) / cos2(x)= 1
What is the second derivative of y equals 2sinx cosx?
y = 2sin(x)cos(x)Use the product rule: uv' + vu' where u is 2sin(x) and v is cos(x) to find first derivative:y' = 2sin(x)(-sin(x)) + cos(x)2cos(x)Simplify:y' = 2cos2(x)-2sin2(x)y' = 2(cos2(x)-sin2(x))Use trig identity cos(2x) = cos2(x)-sin2(x):y' = 2cos(2x)Take second derivative using chain rule:y'' = 2(-sin(2x)cos(2x))Simplify:y'' = -2sin(2x)(2)Simplify:y'' = -4sin(2x)y'' = -4sin(2x)
How does 1-sin squared x times 1 plus tan suared x equals 1?
Use these identities: sin2(x) + cos2(x) = 1, and tan(x) = sin(x)/cos(x) For clarity, the functions are written here without their arguments (the "of x" part). (1 - sin2) = cos2 (1 + tan2) = (1 + sin2/cos2) = (cos2+sin2) / cos2 = 1/cos2 Multiply them: (cos2) times (1/cos2) = 1'QED'
