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Sin^(2)[X] = 1 - Cos^(2)[X]
It is based on Pythagorean theorem .
Algebraically rearrange
Sin^(2)[x] + Cos^(2)[X[ = 1^(2) = 1
Note how it looks like the Pythagorean triangle h^(2) = a^(2) + b^(2) .
What else can I help you with?
What is sin2 equal to?
The expression (\sin^2(x)) (read as "sine squared of x") is equal to ((\sin(x))^2), meaning it represents the square of the sine of the angle (x). This value can vary between 0 and 1, depending on the angle (x). Additionally, it can be expressed using the Pythagorean identity as (\sin^2(x) = 1 - \cos^2(x)).
What does negative sine squared plus cosine squared equal?
To determine what negative sine squared plus cosine squared is equal to, start with the primary trigonometric identity, which is based on the pythagorean theorem...sin2(theta) + cos2(theta) = 1... and then solve for the question...cos2(theta) = 1 - sin2(theta)2 cos2(theta) = 1 - sin2(theta) + cos2(theta)2 cos2(theta) - 1 = - sin2(theta) + cos2(theta)
In a right angled triangle ABC the hypotenuse AC is equal to what?
In a right angled triangle its hypotenuse when squared is equal to the sum of its squared sides which is Pythagoras' theorem for a right angle triangle.
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) ).
If a cos theta plus b sin theta equals 8 and a sin theta - b cos theta equals 5 show that a squared plus b squared equals 89?
There is a hint to how to solve this in what is required to be shown: a and b are both squared.Ifa cos θ + b sin θ = 8a sin θ - b cos θ = 5then square both sides of each to get:a² cos² θ + 2ab cos θ sin θ + b² sin² θ = 64a² sin² θ - 2ab sin θ cos θ + b² cos² θ = 25Now add the two together:a² cos² θ + a² sin² θ + b² sin² θ + b² cos² θ = 89→ a²(cos² θ + sin² θ) + b² (sin² θ + cos² θ) = 89using cos² θ + sin² θ = 1→ a² + b² = 89
What does 2sin squared theta equal and why?
It is 2*sin(theta)*sin(theta) because that is how multiplication is defined!
Does cos squared x over sin squared x equal cot squared x?
Yes. Except where sin x = 0, because then you would be dividing by zero so the quotient is undefined.
What does cos squared x - Sin squared x equal?
2 x cosine squared x -1 which also equals cos (2x)
What is Cos squared x equal to?
Cos^2 x = 1 - sin^2 x
What is sine squared?
Answer 1 Put simply, sine squared is sinX x sinX. However, sine is a function, so the real question must be 'what is sinx squared' or 'what is sin squared x': 'Sin(x) squared' would be sin(x^2), i.e. the 'x' is squared before performing the function sin. 'Sin squared x' would be sin^2(x) i.e. sin squared times sin squared: sin(x) x sin(x). This can also be written as (sinx)^2 but means exactly the same. Answer 2 Sine squared is sin^2(x). If the power was placed like this sin(x)^2, then the X is what is being squared. If it's sin^2(x) it's telling you they want sin(x) times sin(x).
What is sin squared x minus cos squared x?
Sin squared, cos squared...you removed the x in the equation.
Does sin squared x plus cos squared x equal 1?
Yes. 'sin2x + cos2x = 1' is one of the most basic identities in trigonometry.
What does 1 - cos squared equal?
Note that an angle should always be specified - for example, 1 - cos square x. Due to the Pythagorean formula, this can be simplified as sin square x. Note that sin square x is a shortcut of (sin x) squared.
What is sin squared x equal to?
sin x times sin x. or 1/cosec2(x) or 1 - cos2(x) or tan2(x)*cos2(x) etc, etc.
What does Sin squared x - Cos squared x equal?
1 - 2cos2(x) and also 2sin2(x) - 1 Take your choice. Use whichever one is more convenient.
What does Sin squared x Cos squared x equal?
sin2 x = (1/2)(1 - cos 2x) cos2 x = (1/2)(1 + cos 2x) Multiplying both you get (1/4) (1 - cos2 2x) Which is equal to (1/4) (1 - (1/2) (1 + cos 4x) = (1/8) (2 - 1 - cos 4x) = (1/8) (1 - cos 4x) Or If it is the trigonomic function, sin squared x and cosine squared x is equal to one
How do you proven that tan squared x-sin squared tan squared x sin squared x?
To prove the identity ( \tan^2 x - \sin^2(\tan^2 x) \sin^2 x = 0 ), we start with the definitions of tangent and sine. Recall that ( \tan^2 x = \frac{\sin^2 x}{\cos^2 x} ). By substituting this into the equation and simplifying using the Pythagorean identity, we can show that both sides of the equation are equal. Thus, the identity holds true for all ( x ) where the functions are defined.
