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Question: How do you find the exact value of \(\sin {85^ \circ }\) using the sum and difference, double angle ...

How do you find the exact value of sin⁡85∘\sin {85^ \circ } using the sum and difference, double angle or half angle formulas?

Explanation

Solution

For solving this particular question , you have to simplify the given expression by reordering the equation , applying trigonometric identity , taking square root. We cannot find the exact value of sin85sin85 using the sum and difference , double angle, or half angle formulas. While solving for this we get quadratic expressions with complex roots. We have an approximate value that is 0.996194690.99619469.

Complete step by step solution:
We know that ,
cos⁡(6x)=32cos⁡6(x)−48cos⁡4(x)+18cos⁡2(x)−1 ⇒cos⁡(6⋅85)=cos⁡(510)=cos⁡(150)=−32 =32cos⁡6(85)−48cos⁡4(85)+18cos⁡2(85)−1  \cos (6x) = 32{\cos ^6}(x) - 48{\cos ^4}(x) + 18{\cos ^2}(x) - 1 \\\ \Rightarrow \cos (6 \cdot 85) = \cos (510) = \cos (150) = - \dfrac{{\sqrt 3 }}{2} \\\ = 32{\cos ^6}(85) - 48{\cos ^4}(85) + 18{\cos ^2}(85) - 1 \\\
Put cos⁡(85)=y\cos (85) = y , we will get the following ,
32y6−48y4+18y2+32−1=032{y^6} - 48{y^4} + 18{y^2} + \dfrac{{\sqrt 3 }}{2} - 1 = 0
For more simplification, put y2=z{y^2} = z ,
32z3−48z2+18z+32−1=032{z^3} - 48{z^2} + 18z + \dfrac{{\sqrt 3 }}{2} - 1 = 0
Now, if you are able to find zz , then you can easily get yyand then finally you will get the value of sin⁡85\sin 85 by using the identity that is sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1 .

Additional Information:
In arithmetic, pure mathematics identities are equalities that involve pure mathematics functions and are true for every worth of the occurring variables that every aspect of the equality is outlined. Geometrically, these are identities involving sure functions of one or additional angles. We have a trigonometry formula which says sin⁡θ=ph\sin \theta = \dfrac{p}{h} , where pp represent length of perpendicular side and hhrepresent length of hypotenuse side. We have another trigonometry formula which says cos⁡θ=bh\cos \theta = \dfrac{b}{h} , where bb represents length of base and hh represent length of hypotenuse side.

Note: Identities are helpful whenever expressions involving pure mathematics functions should be simplified. An important application is that the combination of non-trigonometric functions: a typical technique involves initial mistreatment the substitution rule with a mathematical relation, then simplifying the ensuing integral with a pure mathematics identity.