Math

QuestionFind the value of tanπ8cot3π8secπ8csc3π8\tan \frac{\pi}{8} \cot \frac{3 \pi}{8} - \sec \frac{\pi}{8} \csc \frac{3 \pi}{8}.

Studdy Solution

STEP 1

Assumptions1. We are working in radians, not degrees. . We know the following trigonometric identities - tan(θ)=sin(θ)cos(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} - cot(θ)=cos(θ)sin(θ)\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} - sec(θ)=1cos(θ)\sec(\theta) = \frac{1}{\cos(\theta)} - csc(θ)=1sin(θ)\csc(\theta) = \frac{1}{\sin(\theta)}

STEP 2

We start by rewriting the given expression using the above identities.
tanπ8cotπ8secπ8cscπ8=sin(π8)cos(π8)cos(π8)sin(π8)1cos(π8)1sin(π8)\tan \frac{\pi}{8} \cot \frac{ \pi}{8}-\sec \frac{\pi}{8} \csc \frac{ \pi}{8} = \frac{\sin(\frac{\pi}{8})}{\cos(\frac{\pi}{8})} \cdot \frac{\cos(\frac{\pi}{8})}{\sin(\frac{\pi}{8})} - \frac{1}{\cos(\frac{\pi}{8})} \cdot \frac{1}{\sin(\frac{\pi}{8})}

STEP 3

We can simplify the above expression by cancelling out the common terms.
sin(π8)cos(π8)cos(3π8)sin(3π8)1cos(π8)1sin(3π8)=cos(3π8)cos(π8)1cos(π8)sin(3π8)\frac{\sin(\frac{\pi}{8})}{\cos(\frac{\pi}{8})} \cdot \frac{\cos(\frac{3\pi}{8})}{\sin(\frac{3\pi}{8})} - \frac{1}{\cos(\frac{\pi}{8})} \cdot \frac{1}{\sin(\frac{3\pi}{8})} = \frac{\cos(\frac{3\pi}{8})}{\cos(\frac{\pi}{8})} - \frac{1}{\cos(\frac{\pi}{8})\sin(\frac{3\pi}{8})}

STEP 4

We can rewrite cos(3π8)\cos(\frac{3\pi}{8}) as cos(π8+π4)\cos(\frac{\pi}{8} + \frac{\pi}{4}) using the identity cos(a+b)=cos(a)cos(b)sin(a)sin(b)\cos(a + b) = \cos(a)\cos(b) - \sin(a)\sin(b).
cos(3π8)cos(π8)1cos(π8)sin(3π8)=cos(π8)cos(π4)sin(π8)sin(π4)cos(π8)1cos(π8)sin(3π8)\frac{\cos(\frac{3\pi}{8})}{\cos(\frac{\pi}{8})} - \frac{1}{\cos(\frac{\pi}{8})\sin(\frac{3\pi}{8})} = \frac{\cos(\frac{\pi}{8})\cos(\frac{\pi}{4}) - \sin(\frac{\pi}{8})\sin(\frac{\pi}{4})}{\cos(\frac{\pi}{8})} - \frac{1}{\cos(\frac{\pi}{8})\sin(\frac{3\pi}{8})}

STEP 5

We know that cos(π4)=sin(π4)=22\cos(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}. Substituting these values into the expression, we getcos(π8)22sin(π8)22cos(π8)1cos(π8)sin(3π8)=22(cos(π8)sin(π8)cos(π8))1cos(π8)sin(3π8)\frac{\cos(\frac{\pi}{8})\frac{\sqrt{2}}{2} - \sin(\frac{\pi}{8})\frac{\sqrt{2}}{2}}{\cos(\frac{\pi}{8})} - \frac{1}{\cos(\frac{\pi}{8})\sin(\frac{3\pi}{8})} = \frac{\sqrt{2}}{2}(\frac{\cos(\frac{\pi}{8}) - \sin(\frac{\pi}{8})}{\cos(\frac{\pi}{8})}) - \frac{1}{\cos(\frac{\pi}{8})\sin(\frac{3\pi}{8})}

STEP 6

We can rewrite sin(3π8)\sin(\frac{3\pi}{8}) as sin(π8+π4)\sin(\frac{\pi}{8} + \frac{\pi}{4}) using the identity sin(a+b)=sin(a)cos(b)+cos(a)sin(b)\sin(a + b) = \sin(a)\cos(b) + \cos(a)\sin(b).
22(cos(π8)sin(π8)cos(π8))1cos(π8)sin(3π8)=22(cos(π8)sin(π8)cos(π8))1cos(π8)(sin(π8)cos(π4)+cos(π8)sin(π4))\frac{\sqrt{2}}{2}(\frac{\cos(\frac{\pi}{8}) - \sin(\frac{\pi}{8})}{\cos(\frac{\pi}{8})}) - \frac{1}{\cos(\frac{\pi}{8})\sin(\frac{3\pi}{8})} = \frac{\sqrt{2}}{2}(\frac{\cos(\frac{\pi}{8}) - \sin(\frac{\pi}{8})}{\cos(\frac{\pi}{8})}) - \frac{1}{\cos(\frac{\pi}{8})(\sin(\frac{\pi}{8})\cos(\frac{\pi}{4}) + \cos(\frac{\pi}{8})\sin(\frac{\pi}{4}))}

STEP 7

Substituting cos(π4)=sin(π4)=22\cos(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} into the expression, we get22(cos(π)sin(π)cos(π))1cos(π)(sin(π)22+cos(π)22)=22(cos(π)sin(π)cos(π))22(cos(π)+sin(π))\frac{\sqrt{2}}{2}(\frac{\cos(\frac{\pi}{}) - \sin(\frac{\pi}{})}{\cos(\frac{\pi}{})}) - \frac{1}{\cos(\frac{\pi}{})(\sin(\frac{\pi}{})\frac{\sqrt{2}}{2} + \cos(\frac{\pi}{})\frac{\sqrt{2}}{2})} = \frac{\sqrt{2}}{2}(\frac{\cos(\frac{\pi}{}) - \sin(\frac{\pi}{})}{\cos(\frac{\pi}{})}) - \frac{2}{\sqrt{2}(\cos(\frac{\pi}{}) + \sin(\frac{\pi}{}))}

STEP 8

Finally, we can simplify the above expression to get the exact value of the given expression.
22(cos(π8)sin(π8)cos(π8))22(cos(π8)+sin(π8))=22=0\frac{\sqrt{2}}{2}(\frac{\cos(\frac{\pi}{8}) - \sin(\frac{\pi}{8})}{\cos(\frac{\pi}{8})}) - \frac{2}{\sqrt{2}(\cos(\frac{\pi}{8}) + \sin(\frac{\pi}{8}))} = \sqrt{2} - \sqrt{2} =0So, the exact value of tanπ8cot3π8secπ8csc3π8\tan \frac{\pi}{8} \cot \frac{3 \pi}{8}-\sec \frac{\pi}{8} \csc \frac{3 \pi}{8} is0.

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