use double-angle recipe to uncover exact values usage double-angle formulas to verify identities use reduction recipe to leveling an expression use half-angle recipe to uncover exact values

Bicycle ramps made for competition (see number (PageIndex1)) must vary in height relying on the skill level the the competitors. For progressed competitors, the angle developed by the ramp and the ground have to be ( heta) such that ( an  heta=dfrac53). The edge is split in half for novices. What is the steepness that the ramp for novices? In this section, we will investigate three extr categories that identities that we deserve to use to answer inquiries such together this one.

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Figure (PageIndex1): bike ramps for advanced riders have actually a steeper incline 보다 those designed for novices.

DOUBLE-ANGLE FORMULAS

The double-angle formulas space summarized as follows:

<eginalign sin(2 heta)&= 2 sin heta cos heta\<4pt> cos(2 heta)&= cos^2 heta-sin^2 heta =1-2 sin^2 heta = 2cos^2 heta-1\<4pt> an(2 heta)&= dfrac2 an heta1- an^2 heta endalign>


How to: offered the tangent of an angle and the quadrant in which it is located, usage the double-angle formulas to discover the specific value

attract a triangle come reflect the offered information. Identify the exactly double-angle formula. Instead of values into the formula based upon the triangle. Simplify.

Example (PageIndex1): using a Double-Angle Formula to find the exact Value entailing Tangent

Given that ( an  heta=−dfrac34) and ( heta) is in quadrant II, find the following:

(sin(2 heta)) (cos(2 heta)) ( an(2 heta))

Solution

If we attract a triangle come reflect the details given, we can find the values essential to deal with the troubles on the image. Us are given ( an  heta=−dfrac34),such that ( heta) is in quadrant II. The tangent of an edge is same to opposing side over the surrounding side, and because ( heta) is in the second quadrant, the nearby side is ~ above the x-axis and also is negative. Use the Pythagorean Theorem to uncover the length of the hypotenuse:

<eginalign* (-4)^2+(3)^2&= c^2\<4pt> 16+9&= c^2\<4pt> 25&= c^2\<4pt> c&= 5 endalign*>

Now we can attract a triangle similar to the one displayed in number (PageIndex2).

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api/deki/files/12456/fig_9.4.3.jpg?revision=1" />Figure (PageIndex3) before we start, we must remember that if (α) is in quadrant III, then (180° to find (cos dfracalpha2),we will write the half-angle formula for cosine, instead of the worth of the cosine we uncovered from the triangle in figure (PageIndex3), and also simplify. <eginalign* cos dfracalpha2&= pm sqrtdfrac1+cos alpha2\<4pt> &= pm sqrtdfrac1+left(-dfrac1517 ight)2\<4pt> &= pm sqrtdfracdfrac2172\<4pt> &= pm sqrtdfrac217cdot dfrac12\<4pt> &= pm sqrtdfrac117\<4pt> &= -dfracsqrt1717 endalign*> We choose the negative value of (cos dfracalpha2) because the edge is in quadrant II because cosine is negative in quadrant II. To find ( an dfracalpha2),we compose the half-angle formula because that tangent. Again, we substitute the worth of the cosine we discovered from the triangle in number (PageIndex3) and also simplify. <eginalign* an dfracalpha2&= pm sqrtdfrac1-cos alpha1+cos alpha\<4pt> &= pm sqrtdfrac1-left(-dfrac1517 ight)1+left(-dfrac1517 ight)\<4pt> &= pm sqrtdfracdfrac3217dfrac217\<4pt> &= pm sqrtdfrac322\<4pt> &= -sqrt16\<4pt> &= -4 endalign*> We pick the an unfavorable value of ( an dfracalpha2) because (dfracalpha2) lies in quadrant II, and tangent is an unfavorable in quadrant II.

Exercise (PageIndex5)

Given that (sin alpha=−dfrac45) and (alpha) lies in quadrant IV, find the specific value of (cos left(dfracalpha2 ight)).​​​​​

prize

(-dfrac2sqrt5)


Example (PageIndex9): recognize the measurement of a fifty percent Angle

Now, us will go back to the difficulty posed in ~ the start of the section. A bike ramp is constructed for high-level competition through an angle of (θ) formed by the ramp and the ground. Another ramp is to it is in constructed half as steep for novice competition. If (tan θ=53) for higher-level competition, what is the measure up of the angle because that novice competition?

Solution

Since the angle for novice vain measures half the steepness of the angle because that the high level competition, and ( an  heta=dfrac53) for high competition, we can find (cos  heta) from the best triangle and the Pythagorean to organize so that we have the right to use the half-angle identities. See figure (PageIndex4).

<eginalign* 3^2+5^2&=34\<4pt> c&=sqrt34 endalign*>

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Figure (PageIndex4)

We watch that (cos  heta=dfrac3sqrt34=dfrac3sqrt3434). We deserve to use the half-angle formula because that tangent: ( an dfrac heta2=sqrtdfrac1−cos  heta1+cos  heta). Since ( an  heta) is in the an initial quadrant, for this reason is ( an dfrac heta2). 

<eginalign* an dfrac heta2&= sqrtdfrac1-dfrac3sqrt34341+dfrac3sqrt3434\<4pt>&= sqrtdfracdfrac34-3sqrt3434dfrac34+3sqrt3434\<4pt>&= sqrtdfrac34-3sqrt3434+3sqrt34\<4pt>&approx 0.57endalign*>

We can take the train station tangent to find the angle: ( an^−1(0.57)≈29.7°). So the edge of the ramp for novice competition is (≈29.7°).

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