5.3 Determine, without the use of a calculator, the value of: \( \sin \left(-200^{\circ}\right) \cdot \cos 310^{\circ}+\tan \left(-135^{\circ}\right) \cdot \cos 380^{\circ} \cdot \sin 230^{\circ} \) 5.4 Prove the following identity: \( \sin 2 \theta+\cos \left(2 \theta-90^{\circ}\right)=4 \sin \theta \cos \theta \)

**5.3** We want to evaluate \[ \sin\left(-200^\circ\right) \cdot \cos 310^\circ + \tan\left(-135^\circ\right) \cdot \cos 380^\circ \cdot \sin 230^\circ. \] 1. **Simplify the first term \(\sin(-200^\circ) \cos 310^\circ\):** - Use the property \(\sin(-\theta)=-\sin\theta\): \[ \sin(-200^\circ)=-\sin(200^\circ). \] - Express \(200^\circ\) as \(180^\circ+20^\circ\) so that \[ \sin(200^\circ)=\sin(180^\circ+20^\circ)=-\sin(20^\circ). \] Thus, \[ \sin(-200^\circ)=-\big[-\sin(20^\circ)\big]=\sin(20^\circ). \] - For \(\cos 310^\circ\): Express \(310^\circ\) as \(360^\circ-50^\circ\) and use the fact that cosine is even: \[ \cos 310^\circ=\cos(50^\circ). \] - Therefore, the first term becomes: \[ \sin(20^\circ) \cos(50^\circ). \] 2. **Simplify the second term \(\tan(-135^\circ)\cos 380^\circ\sin 230^\circ\):** - Use \(\tan(-\theta)=-\tan\theta\): \[ \tan(-135^\circ)=-\tan(135^\circ). \] - Notice that \(135^\circ=180^\circ-45^\circ\), and using the identity \(\tan(180^\circ-\theta)=-\tan\theta\): \[ \tan(135^\circ)=-\tan(45^\circ)=-1. \] Thus, \[ \tan(-135^\circ)=-(-1)=1. \] - Next, reduce \(\cos380^\circ\): since \(380^\circ=360^\circ+20^\circ\), we have \[ \cos380^\circ=\cos20^\circ. \] - For \(\sin230^\circ\): write \(230^\circ=180^\circ+50^\circ\), and use the property \(\sin(180^\circ+\theta)=-\sin\theta\): \[ \sin230^\circ=-\sin50^\circ. \] - Therefore, the second term becomes: \[ 1\cdot \cos20^\circ\cdot \left(-\sin50^\circ\right)=-\cos20^\circ\sin50^\circ. \] 3. **Combine the two results:** \[ \sin(20^\circ)\cos(50^\circ)-\cos(20^\circ)\sin(50^\circ). \] Recognize the sine difference formula: \[ \sin A\cos B-\cos A\sin B=\sin(A-B). \] Let \(A=20^\circ\) and \(B=50^\circ\), then \[ \sin(20^\circ)\cos(50^\circ)-\cos(20^\circ)\sin(50^\circ)=\sin(20^\circ-50^\circ)=\sin(-30^\circ). \] And since \(\sin(-\theta)=-\sin\theta\): \[ \sin(-30^\circ)=-\sin30^\circ=-\frac{1}{2}. \] Thus, the value of the expression is \[ -\frac{1}{2}. \] --- **5.4** We need to prove the identity \[ \sin 2\theta+\cos\left(2\theta-90^\circ\right)=4\sin\theta\cos\theta. \] 1. **Express \(\cos(2\theta-90^\circ)\) in terms of sine:** Recall the co-function identity: \[ \cos\left(x-90^\circ\right)=\sin x. \] Setting \(x=2\theta\) gives: \[ \cos(2\theta-90^\circ)=\sin2\theta. \] 2. **Substitute back into the left-hand side of the identity:** \[ \sin2\theta+\cos(2\theta-90^\circ)=\sin2\theta+\sin2\theta=2\sin2\theta. \] 3. **Express \(\sin2\theta\) in terms of \(\sin\theta\) and \(\cos\theta\):** Using the double-angle formula: \[ \sin2\theta=2\sin\theta\cos\theta. \] Then, \[ 2\sin2\theta=2\cdot\big(2\sin\theta\cos\theta\big)=4\sin\theta\cos\theta. \] Thus, we have shown that \[ \sin 2\theta+\cos(2\theta-90^\circ)=4\sin\theta\cos\theta, \] which completes the proof.