Question

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Answer 1

Using the sum and difference formulae respectively, the values of  $\frac{13\pi }{12}$ include (a) tan∅=0.26794919 and (b) cos∅= -2.7321

How to calculate use the sum and difference formulae?

Trigonometry is defined as the study of the properties of triangles and trigonometric functions and the applications of the trigonometric ratios of sine, cosine and tangent in solving problems.

The given value is  $\frac{13\pi }{12}$

(a) Using the sum formula Tan(u+v) =$\frac{tanu +tanv}{1tanv-tanu}$

Let us split the value into two angles.

tan[$\frac{3\pi }{4} +\frac{\pi x}{3}]=tan\frac{3\pi }{4} +tan\frac{\pi x}{3}$ ] ÷1-tan$\frac{3\pi }{4} tan\frac{\pi x}{3}$

The equivalent trig. from the first quadrant is negative because tangent is in the second quadrant.

⇒ $\frac{-1+\sqrt{3} }{1-(-1)\sqrt{3} }$

⇒ $\frac-1+\sqrt{3} }{1+\sqrt{3} }$

Rationalize the denominator we have

$\frac{-1+\sqrt{3} }{1+\sqrt{3} } * \frac{1-\sqrt{3} }{1-\sqrt{3} }$

simplifying the surd we have -2+$\sqrt{3}$

When this is expressed in decimal, we have 0.29794919

(b) The cosine difference formula is Cos(A-B)=cosACosB+sinASinB

The given value is  $\frac{13\pi }{12}$

⇒ cos$cos\frac{3\pi }{4} cos\frac{\pi }{3} + sin\frac{3\pi }{4} sin\frac{\pi }{3}$

This implies that [(-1+$\sqrt{3}$)(1-(-1)$\sqrt{3}$] +[9-1+$\sqrt{3)} 1+(-1)\sqrt{3}$

⇒ [(-1+$\sqrt{3})(\sqrt{3} ) +(-1+\sqrt{3} )(1-\sqrt{3} )$

Simplifying the surd we have. -1-$\sqrt{3}$

-1-1.7321

When this is expressed in decimal, we have -2.7321

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