09 Section 7. Double and Half Angle Fmulas To derive the double-angles fmulas, we will use the sum of two angles fmulas that we developed in the last section. We will let α θ and β θ: cos(θ) cos(θ + θ) cos(θ)cos(θ) sin(θ)sin(θ) cos (θ) sin (θ) sin(θ) sin(θ + θ) sin(θ)cos(θ) + cos(θ)sin(θ) sin(θ)cos(θ) tan(θ) tan(θ + θ) tan(θ)+tan(θ) tan(θ) tan(θ) tan(θ) tan (θ) F the double angle fmula f the cosine function, we can derive several variations of it using the Pythagean Fmula sin (θ) + cos (θ). Solving f cos (θ), we get cos (θ) sin (θ). Thus, cos (θ) sin (θ) ( sin (θ)) sin (θ) sin (θ) Now, solving f sin (θ), we get sin (θ) cos (θ). Hence cos (θ) sin (θ) cos (θ) [ cos (θ)] cos (θ) + cos (θ) cos (θ) Double-Angle Fmulas: ) cos(θ) cos (θ) sin (θ) sin (θ) cos (θ) ) sin(θ) sin(θ)cos(θ) ) tan(θ) tan(θ) tan (θ) Objective : Use Double-angle Fmulas to Find Exact Values. Find the Exact Value of the following: Ex. Given tan(θ), π < θ < π, find a) tan(θ) b) sin(θ) c) cos(θ) a) Plug the value of tan(θ) into our double-angle fmula: tan(θ) tan(θ) tan (θ) ( ( ) ) ( )
b) Befe we can apply the double-angle fmula f the sine function, we need to find sin(θ) and cos(θ). Since θ is in the second quadrant, y will be positive and x will be negative. tan(θ) 0 y, so x and y. Now, find r: x r x + y r ( ) + ( ) r + r r ± (reject the negative answer) Thus, sin(θ) y r cos(θ) x r Now, we can find sin(θ): sin(θ) sin(θ)cos(θ) ( )(. c) Using the results fm part b, we get: cos(θ) cos (θ) sin (θ) (. ) 8 9 ) ( 9 and ) 9 9 9 Objective : Use the double-angle fmulas to establish identities. Establish the following identities: Ex. a cos (θ) sin (θ) cos(θ) Ex. b cos(θ) +sin(θ) cot(θ) cot(θ)+ Ex. c cot(θ) (cot(θ) tan(θ)) a) We will start with the left side: cos (θ) sin (θ) (fact) [cos (θ) sin (θ)][cos (θ) + sin (θ)] (cos (θ) + sin (θ) ) [cos (θ) sin (θ)][] cos (θ) sin (θ) (apply the double-angle identity) cos(θ)
b) We will start with the right side; cot(θ) cot(θ)+ cos(θ ) sin(θ ) cos(θ ) sin(θ ) + (write in terms of sine and cosine) (multiply the top & bottom by sin(θ)) cos(θ ) sin(θ ) sin(θ) sin(θ) cos(θ ) sin(θ ) sin(θ)+ sin(θ) cos(θ) sin(θ) cos(θ) +sin(θ) If we examine the left side f a moment, we have cos(θ) in the numerat which is equal to cos (θ) sin (θ). This suggests that we need to multiply top and bottom by the conjugate of the numerat: cos(θ) sin(θ) cos(θ)+sin(θ) cos(θ)+sin(θ) (expand) cos(θ) +sin(θ) cos (θ) sin (θ) cos (θ)+cos(θ)sin(θ)+sin (θ) cos (θ) sin (θ) cos (θ)+sin (θ)+ cos(θ)sin(θ) cos (θ) sin (θ) +cos(θ) sin(θ) cos(θ) +sin(θ) c) Let's start with the left side: cot(θ) tan(θ) tan(θ) tan (θ) tan (θ) tan(θ) tan(θ ) tan (θ) tan(θ ) tan(θ) tan(θ ) tan(θ ) tan(θ) (cot(θ) tan(θ)) (regroup the denominat) (cos (θ) + sin (θ) ) (apply the double-angle fmulas) The identity has been established (apply double-angle fmula f tangent) (invert and multiply) (multiply top and bottom by (cot(θ) tan(θ) ) tan(θ) )
Solve f all values in [0, π): Ex. sin(θ)sin(θ) cos(θ) sin(θ)sin(θ) cos(θ) (get zero on one side) sin(θ)sin(θ) cos(θ) 0 (sin(θ) sin(θ)cos(θ)) sin(θ)cos(θ)sin(θ) cos(θ) 0 (fact out cos(θ)) cos(θ)[sin (θ) ] 0 (solve) cos(θ) 0 sin (θ) 0 Ex. θ π π sin (θ) sin(θ) ± ± ± The reference angle is π, so the four angles are π, π π π, π + π 5π, π π 7π So, the solution is { π, π, π, 5π, π, 7π }. cos(θ) sin (θ) cos(θ) sin (θ) (cos(θ) sin (θ)) sin (θ) sin (θ) (add sin (θ) to both sides) There is no solution. Ex. 5 tan(θ) + cos(θ) 0 The tangent function is undefined when the cos(θ) 0 θ π. Writing the general fm of the solution, our restriction is θ π π + kπ + kπ (solve f θ) θ π π + kπ + kπ, k is an integer k 0 θ π + (0)π π π π + (0)π π k θ π 5π π + ()π Thus, our restriction is θ π, π, 5π 7π,. + ()π 7π
tan(θ) + cos(θ) 0 sin(θ) cos(θ) + cos(θ) 0 sin(θ)cos(θ) sin (θ) sin(θ)cos(θ) sin (θ) (tan(θ) sin(θ) cos(θ) ) (use the double angle identities) + cos(θ) 0 (multiply by sin (θ)) ( sin (θ)) + cos(θ) ( sin (θ)) 0 ( sin (θ)) sin(θ)cos(θ) + cos(θ)( sin (θ)) 0 (fact out cos(θ)) cos(θ)[sin(θ) + ( sin (θ)] 0 (simplify inside the bracket) cos(θ)[ sin (θ) + sin(θ) + ] 0 (fact out ) cos(θ)[sin (θ) sin(θ) ] 0 (x x (x + )(x ) cos(θ)[sin(θ)+ ][sin(θ) ] 0 (solve) cos(θ) 0, sin(θ) + 0, sin(θ) 0 cos(θ) 0, sin(θ), sin(θ) θ π π, θ π + π 7π π π π, θ π None of these values match our restrictions, so the solution is { π, 7π, π, π }. We will first derive the half angle fmula f the sine function. Recall that cos(θ) sin (θ). We will now solve this f sin(θ): sin (θ) cos(θ) (subtract ) sin (θ) cos(θ) (divide by ) sin (θ) cos(θ) (use the square root property) sin(θ) ± sin( α ) ± cos(θ) cos(α) (Let α Now, we will derive the half angle fmula f cos(θ): cos (θ) cos(θ) (add ) cos (θ) + cos(θ) (divide by ) cos (θ) +cos(θ) cos(θ) ± cos( α ) ± +cos(θ) θ, then α θ) (use the square root property) (Let α θ, then α θ)
F the half angle fmula f the tangent function, we will use the quotient fmula: tan ( α ) sin ( α ) sin ( α cos ( α ) ) cos ( α ) cos(α) cos(α) So, tan ( α ) cos(α) tan( α ) ± cos(α). cos(α) (use the square root property) Sometimes, it might be better to use the fmula you get befe you use the square root property. Half angle fmulas sin( α ) ± cos( α ) ± cos(α) sin ( α ) cos(α) cos ( α ) tan( α ) ± cos(α) tan ( α ) cos(α) Sign equals the sign of trigonometric function in the quadrant of α/. Find the following: Ex. Write sin (θ) as an equivalent expression with the powers of the sine and cosine equal to one. sin (θ) (write as a perfect square and θ θ ) [sin ( θ )] (apply the nd half-angle fmula f sine) cos(θ) [ ] (expand) cos(θ) cos(θ) cos(θ)+cos (θ) cos(θ) + cos (θ) (write θ θ in the last term) cos(θ) + cos ( θ ) (apply the nd fmula f cosine) cos(θ) + +cos(θ) ( )
5 cos(θ) + 8 + cos(θ) 8 8 cos(θ) + 8 cos(θ) Establish the identity: Ex. 7a tan( α ) cos(α) Ex. 7b tan( α ) a) We will start with the right side. Since sin ( α ) cos(α), then sin ( α ) cos(α). Also, sin[( α )] sin( α )cos( α ). Thus, cos(α) sin ( α ) sin( α )cos( α ) (reduce) sin( α ) cos( α ) tan( α ) The identity has been established. b) tan( α cos(α) cos(α) ) (use the result from part a) (multiply top & bottom by the conjugate of cos(α)) cos (α) () sin (α ) () (expand the numerat) (but, sin (α) cos (α)) (reduce) The identity has been established. The advantage of these two fmulas f the half-angle of tangent is you do not have to wry about determining the sign. Other half-angle fmulas f tangent: tan( α ) cos(α) tan( α ) Now, let's apply these fmulas to an application problem from the book (Sullivan's PreCalculus, 9 th edition, 0, page 9, exercise 95.)
Verify the following: Ex. 8 Given D D W csc(θ) cot(θ) W csc(θ) cot(θ), show that W Dtan( θ ) (multiply by (csc(θ) cot(θ)) (csc(θ) cot(θ))d (csc(θ) cot(θ)) W csc(θ) cot(θ) (csc(θ) cot(θ))d W W D(csc(θ) cot(θ)) (rewrite in terms of sine and cosine) W D( sin(θ) cos(θ) sin(θ) ) W D( cos(θ) sin(θ) W Dtan( θ ) (combine) ) (apply the half-angle fmula) Objective : Use half-angle Fmulas to Find Exact Values Find the exact value of the following: Ex. 9a sin(5 ) Ex. 9b cos(.5 ) a) sin(5 ) sin( 0o ) ± cos(0 o ) Since 5 is in quadrant I, then the sine function is positive: cos(0 o ) (cos(0 ) ) (multiply the top & bottom by ) b) Since the cosine function is even, then cos(.5 ) cos(.5 ). Thus, +cos(5 o ) cos(.5 ) cos( 5o ) ± Since.5 is in quadrant II, then cosine function is negative:
7 +cos(5 o ) (cos(5 ) cos(5 ) +( ) (multiply the top & bottom by ) Ex. 0 Given cos(θ) π < θ < π, find a) sin( θ ) b) cos( θ ) c) tan( θ ) First. let's figure out what quadrant θ lies in: π < θ < π (divide by ) π < θ < π Thus, θ a) sin( θ ) ± cos(θ) Since θ cos(θ) b) cos( θ ) ± +cos(θ) Since θ is in quadrant II. is in quadrant II, then the sine function is positive:. (cos(θ) ) (multiply the top & bottom by ) is in quadrant II, then the cosine function is negative: +cos(θ) (cos(θ) )
8 + + + c) tan( θ ) ± cos(θ) +cos(θ) Since θ. (multiply the top & bottom by ) is in quadrant II, then the tangent function is negative: cos(θ) +cos(θ) + + (cos(θ) ) (multiply the top & bottom by ) Solve f all angles in the interval [0, π): Ex. sin (θ) + cos (θ) 0 sin (θ) + cos (θ) 0 (F + L (F + L)(F FL + L )) (sin(θ) + cos(θ))(sin (θ) sin(θ)cos(θ) + cos (θ)) 0 (sin (θ) + cos (θ) in the nd parenthesis) (sin(θ) + cos(θ))( sin(θ)cos(θ)) 0 (sin(θ) sin(θ)cos(θ)) (sin(θ) + cos(θ))( sin(θ)) 0 (solve) sin(θ) + cos(θ) 0 sin(θ) 0 cos(θ) sin(θ) sin(θ) (square both sides) no solution cos (θ) sin (θ) (subtract sin (θ) from both sides) cos (θ) sin (θ) 0 (cos(θ) cos (θ) sin (θ)) cos(θ) 0 θ π θ π π + kπ π We will need to write our general solution: + kπ (solve f θ)
θ π π + kπ + kπ, k is an integer k 0 θ π + (0)π π π π + (0)π k θ π 5π + ()π π 7π + ()π Since we squared the equation, we have to check f the solutions. θ π θ π θ 5π θ 7π sin ( π ) + cos ( π ) + 0 (reject) sin ( π ) + cos ( π ) + ( ) 0 (keep) sin ( 5π ) + cos ( 5π ) + ( ) 0 (reject) sin ( 7π ) + cos ( 7π ) ( ) + 0 (keep) The solution is { π, 7π } Let' summarize the fmulas that we derived in this section: Double-Angle Fmulas: ) cos(θ) cos (θ) sin (θ) sin (θ) cos (θ) ) sin(θ) sin(θ)cos(θ) ) tan(θ) tan(θ) tan (θ) 9 Half angle fmulas ) sin( α ) ± cos(α) 5) cos( α ) ± ) tan( α ) ± cos(α) b) tan( α ) cos(α) sin ( α ) cos(α) cos ( α ) tan ( α ) cos(α) c) tan( α )