Math  /  Trigonometry

QuestionUse the half-angle formulas to determine the exact values of the sine, cosine, and tangent of the angle. 6730sin(6730)=cos(6730)=tan(6730)=\begin{array}{c} 67^{\circ} 30^{\prime} \\ \sin \left(67^{\circ} 30^{\prime}\right)=\square \\ \cos \left(67^{\circ} 30^{\prime}\right)=\square \\ \tan \left(67^{\circ} 30^{\prime}\right)=\square \end{array}

Studdy Solution
Simplify the expression:
tan(67.5)=1+22122 \tan(67.5^\circ) = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{1 - \frac{\sqrt{2}}{2}}}
tan(67.5)=2+222 \tan(67.5^\circ) = \sqrt{\frac{2 + \sqrt{2}}{2 - \sqrt{2}}}
Rationalize the denominator:
tan(67.5)=(2+2)(2+2)(22)(2+2) \tan(67.5^\circ) = \sqrt{\frac{(2 + \sqrt{2})(2 + \sqrt{2})}{(2 - \sqrt{2})(2 + \sqrt{2})}}
tan(67.5)=(2+2)242 \tan(67.5^\circ) = \sqrt{\frac{(2 + \sqrt{2})^2}{4 - 2}}
tan(67.5)=4+42+22 \tan(67.5^\circ) = \sqrt{\frac{4 + 4\sqrt{2} + 2}{2}}
tan(67.5)=3+22 \tan(67.5^\circ) = \sqrt{3 + 2\sqrt{2}}
The exact values are:
sin(6730)=2+22 \sin(67^\circ 30') = \frac{\sqrt{2 + \sqrt{2}}}{2} cos(6730)=222 \cos(67^\circ 30') = \frac{\sqrt{2 - \sqrt{2}}}{2} tan(6730)=3+22 \tan(67^\circ 30') = \sqrt{3 + 2\sqrt{2}}

View Full Solution - Free
Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord