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Find the value of:(i) $\sin 75^{\circ}$(ii) $\tan 15^{\circ}$

Precalculus

Chapter 3

Trigonometric Functions

Section 3

Trigonometry

Piedmont College

University of Michigan - Ann Arbor

Boston College

Lectures

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in this kitchen we have to find the value of sane 75 degree. So here it can be written as sign 45 degree blows 30 degree. Now here we have a formula for this sane A plus B equal to sign A. Times goes B plus goals A. Times sign B. So here you can see that equal to 45 degree and be equal to 30 degree. So now we apply that formula. So here saying 45 degree positive 30 degree equal to sane 45 degree times of course 30 degree plus, of course 45 degree times sine 30 degree. So now we substitute the value. So here we get signed 45 degree well is one upon, it's quiet too. Times goes 30 degree value is square or three upon to plus goes 45 degree value is one upon square or two times the value of signed 30 degree value is one up on two. Offer. Simplify this. So here we get squared of three plus one upon two times it's quite or two. So here it is our answer for part A now we sold part second. So now here now here we have to find the value of then 15 degrees it can be written as then 45 degree negative 30 degrees. So here we have a formula for this which is an E. Negative B equal to N. E. Negative than be upon one positive than eight times than be. Now here you can see that equal to 45 degrees ah be equal to 30 degree. So now we apply this formula. So here we get Ben 45 degrees negative 30 degree equal to and 45 degree negative then 30 degree upon 1 90 then 45 degrees times. Please put here positive one positive 10 45 degree times. Ben 30 degree. Now we substitute the value. So here we get 10 45 degree value is one, negative 10 30 degree value is one upon square or three upon one. Plus 10 45 degree value is one. So we put here times one upon square root three. So now we simplify this and here we get squared of three negative one upon square three positive one. No we do rationalizations. So square root three negative one upon square or three negative one. So now we get squared off square or three negative one upon three negative. Now we get after simplify this four negative two times square or three upon to equal to two times so it to negative square or three. So here you can see that and 15 degrees value is too negative square three. So it is a final answer for part second.

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