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Question: How do you find the exact value of \[cos58\] using the sum and difference, double angle or half angl...

How do you find the exact value of cos58cos58 using the sum and difference, double angle or half angle formulas?

Explanation

Solution

First, we need to analyze the data so that we can solve the problem. Here we are asked to find the exact value of cos58cos58 . Since we are asked to find the exact value, we need to use the sum and difference, double angle, or half-angle formulae. The exact value is the value that we cannot estimate accurately.

Complete answer:
It's the nth Chebyshev Polynomial of the first sort and Tn(x)Tn(x) is one of the roots of T44(x)=T46(x)T44(x) = - T46(x). One of the forty-six roots of is:
Explanation:
5858^\circ is not a multiple of 33^\circ . Multiples 11^\circ that are not multiples of 33^\circ are not constructible using a straightedge and compass, and their trig functions are not the result of some integer composition including addition, subtraction, multiplication, division, and square roots.
That doesn't mean we can't record an expression 5858^\circ . Let us interpret the degree sign as a fact 2π360\dfrac{{2\pi }}{{360}}
ei58=cos58+i  sin58e{i^{{{58}^\circ }}} = cos58^\circ + i\;sin58^\circ
ei58=cos58i  sin58e{i^{ - {{58}^\circ }}} = cos58^\circ - i\;sin58^\circ
(ei58+ei58)=2×cos58({e^{i58^\circ }} + {e^{ - i58^\circ }})= 2 \times {\cos 58^\circ }
cos58=12(ei58+ei58)\cos 58^\circ = \dfrac{1}{2}({e^{i58^\circ }} + {e^{ - i58^\circ }})
That's not very helpful.
We can try to write out a polynomial equation with cos58cos58^\circ as one of its roots, but it will most likely be too large to fit.
θ=2\theta = 2^\circ is 180180 the of the circle. Because, cos88=cos92\cos 88^\circ = - \cos 92^\circ it means, cos2\cos 2^\circ satisfies
cos(44θ)=cos(46θ)\cos (44\theta ) = - \cos (46\theta )
cos(18044θ)=cos(46θ)\cos (180^\circ - 44\theta ) = \cos (46\theta )
Let us solve θ\theta it first.   cosx=cosa\;cosx = cosa has root x=±a+360kx = \pm a + 360^\circ k
kk referred to an integer.
18046θ=±44θ360k180^\circ - 46\theta = \pm 44\theta - 360^\circ k
46θ±44θ=180+360k46\theta \pm 44\theta = 180^\circ + 360^\circ k
θ=2+4k\theta = 2^\circ + 4^\circ k
Or, θ=90+180k\theta = 90^\circ + 180^\circ k
That's a lot of roots, and we see cos=58\cos = 58^\circ within it.
The polynomials Tn(x)Tn(x), also known as the Chebyshev polynomials First-order polynomials fulfill cos(nθ)=Tn(cosθ)cos(n\theta ) = Tn(cos\theta ) and have integer coefficients. The first few are known from the double and triple angle formulas:
cos(0θ)=1    cos(0\theta ) = 1\;\;So, T0(x)=1T0(x) = 1
cos(1θ)=cosθ  cos(1\theta ) = cos\theta \; So, T1(x)=xT1(x) = x
cos(2θ)=2cos2θ1  cos(2\theta ) = 2co{s^2}\theta - 1\; So,   T2(x)=2x21\;T2(x) = 2{x^2} - 1
cos(3θ)=4cos3θ3cosθcos(3\theta ) = 4co{s^3}\theta - 3cos\theta So,   T3(x)=4x43x\;T3(x) = 4{x^4} - 3x
We can verify a suitable recursion relation:
Tn+1(x)=2xTn(x)Tn1(x){T_{n + 1}}(x) = 2x{T_n}(x) - {T_{n - 1}}(x)
So, in theory, we can generate these for as large a nn as we want.
If we let x=cosx = cos, then the equation becomes
cos(44θ)=cos(46θ)cos(44\theta ) = - cos(46\theta )
Gets, T44(x)=T46(x){T_{44}}(x) = - {T_{46}}(x)
I will solve the equation just to see how the math works:
8796093022208{x^{44}}$$$$ - 96757023244288{x^{42}}$$$$ + 495879744126976{x^{40}}$$$$ - 1572301627719680{x^{38}}$$$$ + 3454150138396672{x^{36}}$$$$ - 5579780992794624{x^{34}}$$$$ + 6864598984556544{x^{32}}$$$$ - 6573052309536768{x^{30}}$$$$ + 4964023879598080{x^{28}}

Note:
There are a lot of formulas available to solve the trigonometric problems. Generally, we use the double angle formula to verify the given identity and we prefer the double angle formula to find the exact value of the trigonometric functions. Also, we apply the reduction formula to simplify the given trigonometric expression.