@Valryan14
Solving Trigonometric Functions of Two Angles.
Today, in this article I'm gonna show to you how to solve a Trigonometric Functions of the Two Angles. This might help you in your math subject especially for those who hates Mathematics. 1.) Find the exact value of of sin75° Solution: **sin(75°) = sin(30°+45°)** Using this formula; ***sin(A+B) = sinAcosB + cosAsinB*** Given: sinA = sin30°= ½ cosB = cos45°= √2/2 cosA = cos30°= √3/2 sinB = sin45°= √2/2 sin(30°+45°) = sin30°cos45° + cos30°sin45° sin(30°+45°) = (1/2)(√2/2) + (√3/2)(√2/2) sin(30°+45°) = (√2/4) + (√6/4) sin(75°) = ****(√2 + √6)/4**** 2. Find the exact value of of tan15° Solution: **tan15° = tan(60° -45°)** Using this formula ***tan(A-B) = (tanA -tanB)/(1 +tanAtanB)*** Given: tanA = tan60° = √3 tanB = tan45° = 1 tan(60° -45°) = (√3 -1)/(1 +√3) tan(60° -45°) = [(√3 -1)/(1 +√3)]*[(1-√3)/(1-√3)] -----> (*Rationalize the denominator*) tan(60° -45°) = (2√3 -4)/(-2) tan15° = -****2√3 +2**** 3. If sinA=2/3, find sin(A-90°). Solution: Using Subtraction Formula***; sin(A-B) = sinAcosB - cosAsinB*** Since sinA=2/3 = o/h, *we’re going to solve for “a” using Phythagorean theorem* h^2 = a^2 +o^2 3^2 = a^2 +2^2 a = √5 sin(A-90°) = sinAcos90° - cosAsin90° Given: sinA =2/3 cosA =a/h =√5/3 --------> (*substitute the value of “a” we have solve ealier)* cos90° =0 sin90° =1 . sin(A-90°) = (2/3)*0 – (√5/3) sin(A-90°) = -****√5/3**** 4. Given sinA =4/5 and cosB =5/13, find sin(A+B) +sin(A-B). Solution: Simplify first *sin(A+B) +sin(A-B)* *sin(A+B) +sin(A-B) = (sinAcosB +cosAsinB)+(sinAcosB -cosAsinB)* *sin(A+B) +sin(A-B) = 2sinAcosB* 2sinAcosB 2(4/5)(5/13) (*substitute the given values)* ****= 8/13**** 5. Find the exact value of (tan73° +tan32°)/(1 –tan73°tan32°). Solution: Notice the given expression above is in the form of “(tanA +tanB)/(1 -tanAtanB)” which is equal to tan(A+B) thus, *tan(73°+32°) = (tan73° +tan32°)/(1 –tan73°tan32°).* tan(73°+32°) = tan(105°) =tan(60°+45°) --------> (*we rewrite the given angle as the sum of two special angle)* *tan(60°+45°) = (tan60° +tan45°)/(1 –tan60°tan45°)* *Given:* tan(45°) =1 tan(60°) =√3 tan(60°+45°) = (tan60° +tan45°)/(1 –tan60°tan45°) tan(60°+45°) = (√3 +1)/(1 –√3) tan(60°+45°) = [(√3 +1)/(1 –√3)]*[( 1 +√3)/( 1 +√3)] --------> (*Rationalize the Denominator)* tan(60°+45°) = (2√3 +4)/(-2) . = ****-√3 -2**** **Conclusion** I know Maths really is hard to understand but if you pursue and enjoy it you will get used to it and you can learn more easily.
No comments yet