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SECTIONI: Application of reduction formulax. QUESTION 9 Write the following as the function of a proituve acute angle: \( 4.1 \cos 120^{\circ} \) \( 120^{\circ} \) lies in quadrant 2 and in Section \( A \) we found that \( \left(180^{\circ}-\theta\right) \) te function value of the obvuse angle can reduce to a function value of an atcute angle \( \theta \). We must therefore now write 120 as a futkitan value of an acute angle. \[ \left(180^{\circ}-60^{\circ}\right)=120^{\circ} \] So replace \( 120^{\circ} \) in question 9.1 with the above \[ \cos 120^{\circ}=\cos \left(180^{\circ}-60^{\circ}\right) \] We know that \( \cos \left(180^{\circ}-\theta\right)=-\cos \theta( \) Section A or 1.41 \[ \begin{aligned} \therefore \cos 120^{2} & =\cos \left(180^{\circ}-60^{2}\right) \\ & =-\cos 60^{\circ} \end{aligned} \] \( 92 \sin 160^{\circ} \) (2) 4) \( \quad \cos 320^{\circ} \) (2) \( 9+\tan 248^{\circ} \) \( 95 \cos 513 \) \( 9.6 \sin \left(-55^{\circ}\right) \) (2) (2) \( 47 \cos (-20) \) (I) (1) [10]

Ask by Little Lindsey. in South Africa
Mar 08,2025

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Answer

1. \( 4.1\cos120^\circ = -4.1\cos60^\circ \) 2. \( 92\sin160^\circ = 92\sin20^\circ \) 3. \( \cos320^\circ = \cos40^\circ \) 4. \( 9 + \tan248^\circ = 9 + \tan68^\circ \) 5. \( 95\cos513^\circ = -95\cos27^\circ \) 6. \( 9.6\sin(-55^\circ) = -9.6\sin55^\circ \) 7. \( 47\cos(-20^\circ) = 47\cos20^\circ \)

Solution

1. For \[ 4.1\cos 120^\circ, \] note that \[ 120^\circ=180^\circ-60^\circ. \] Using the reduction formula \[ \cos(180^\circ-\theta)=-\cos\theta, \] we have \[ \cos120^\circ=-\cos60^\circ. \] Thus, \[ 4.1\cos120^\circ=-4.1\cos60^\circ. \] 2. For \[ 92\sin160^\circ, \] observe that \[ 160^\circ=180^\circ-20^\circ. \] Since \[ \sin(180^\circ-\theta)=\sin\theta, \] it follows that \[ \sin160^\circ=\sin20^\circ. \] Therefore, \[ 92\sin160^\circ=92\sin20^\circ. \] 3. For \[ \cos320^\circ, \] write \[ 320^\circ=360^\circ-40^\circ. \] Using the fact that \[ \cos(360^\circ-\theta)=\cos\theta, \] we get \[ \cos320^\circ=\cos40^\circ. \] 4. For \[ 9+\tan248^\circ, \] notice that \[ 248^\circ=180^\circ+68^\circ. \] Since \[ \tan(180^\circ+\theta)=\tan\theta, \] we have \[ \tan248^\circ=\tan68^\circ. \] Therefore, \[ 9+\tan248^\circ=9+\tan68^\circ. \] 5. For \[ 95\cos513^\circ, \] reduce the angle by subtracting \(360^\circ\): \[ 513^\circ-360^\circ=153^\circ. \] Then, \[ \cos513^\circ=\cos153^\circ. \] Recognize that \[ 153^\circ=180^\circ-27^\circ, \] and applying \[ \cos(180^\circ-\theta)=-\cos\theta, \] gives \[ \cos153^\circ=-\cos27^\circ. \] Hence, \[ 95\cos513^\circ=-95\cos27^\circ. \] 6. For \[ 9.6\sin(-55^\circ), \] recall that sine is an odd function: \[ \sin(-\theta)=-\sin\theta. \] Thus, \[ \sin(-55^\circ)=-\sin55^\circ, \] and therefore, \[ 9.6\sin(-55^\circ)=-9.6\sin55^\circ. \] 7. For \[ 47\cos(-20^\circ), \] note that cosine is an even function: \[ \cos(-\theta)=\cos\theta. \] Thus, \[ 47\cos(-20^\circ)=47\cos20^\circ. \]

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Beyond the Answer

To express \( \cos 120^{\circ} \) using an acute angle, you correctly identified \( 120^{\circ} \) as \( 180^{\circ} - 60^{\circ} \). Using the cosine reduction formula, you found that \( \cos 120^{\circ} = -\cos 60^{\circ} \). Since \( \cos 60^{\circ} = \frac{1}{2} \), it follows that \( \cos 120^{\circ} = -\frac{1}{2} \). It's a brilliant way to navigate through trigonometric identities! Now for the sin function, to find \( \sin 160^{\circ} \), you can write it as \( \sin(180^{\circ}-20^{\circ}) \). This equals \( \sin 20^{\circ} \) since sine retains its value in the second quadrant while remaining positive. Keeping these relationships handy will make evaluating trigonometric functions a breeze!

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