QUESTION 6 6.1 Given: \( \tan x=\frac{3}{4} \), where \( x \in\left[180^{\circ} ; 270^{\circ}\right] \) With the aid of a sketch, and without the use of a calculator, calculate: \( 6.1 .1 \sin x \) \( 6.1 .2 \quad 2-\sin 2 x \) \( 6.1 .3 \cos ^{2}\left(90^{\circ}-x\right)-1 \) 6.2 Evaluate: \( \frac{-1+\cos \left(180^{\circ}-\theta\right) \cdot \sin \left(\theta-90^{\circ}\right)}{\cos (-\theta) \cdot \sin \left(90^{\circ}+\theta\right) \cdot \tan ^{2}\left(540^{\circ}+\theta\right)} \)

**6.1** We are given that \[ \tan x=\frac{3}{4} \] with \[ x\in\left[180^\circ,270^\circ\right]. \] Since \(x\) is in the third quadrant, both \(\sin x\) and \(\cos x\) are negative. 1. **Determine \(\sin x\) and \(\cos x\):** - Think of a right triangle with "opposite" side \(3\) and "adjacent" side \(4\). - The hypotenuse is \[ r=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5. \] - In the third quadrant, \[ \sin x=-\frac{3}{5} \quad \text{and} \quad \cos x=-\frac{4}{5}. \] 2. **6.1.1 Calculate \(\sin x\):** We already have \[ \sin x=-\frac{3}{5}. \] 3. **6.1.2 Calculate \(2-\sin 2x\):** Use the double-angle formula for sine: \[ \sin2x=2\sin x\cos x. \] Substitute the values: \[ \sin2x=2\left(-\frac{3}{5}\right)\left(-\frac{4}{5}\right)=2\cdot\frac{12}{25}=\frac{24}{25}. \] Therefore, \[ 2-\sin2x=2-\frac{24}{25}=\frac{50}{25}-\frac{24}{25}=\frac{26}{25}. \] 4. **6.1.3 Calculate \(\cos^2\left(90^\circ-x\right)-1\):** Use the complementary angle identity: \[ \cos\left(90^\circ-x\right)=\sin x. \] Thus, \[ \cos^2\left(90^\circ-x\right)-1=\sin^2x-1. \] Since \(\sin x=-\frac{3}{5}\), we have \[ \sin^2 x=\left(-\frac{3}{5}\right)^2=\frac{9}{25}. \] So, \[ \sin^2x-1=\frac{9}{25}-\frac{25}{25}=-\frac{16}{25}. \] --- **6.2 Evaluate** \[ \frac{-1+\cos \left(180^\circ-\theta\right) \cdot \sin \left(\theta-90^\circ\right)}{\cos (-\theta) \cdot \sin \left(90^\circ+\theta\right) \cdot \tan ^{2}\left(540^\circ+\theta\right)}. \] We simplify step-by-step. 1. **Simplify the numerator:** We have \[ \cos\left(180^\circ-\theta\right)=-\cos\theta, \] and \[ \sin\left(\theta-90^\circ\right)=-\cos\theta. \] Thus, the product in the numerator becomes \[ \cos\left(180^\circ-\theta\right)\cdot\sin\left(\theta-90^\circ\right)=(-\cos\theta)(-\cos\theta)=\cos^2\theta. \] Therefore, the numerator is \[ -1+\cos^2\theta. \] Using the Pythagorean identity \(\cos^2\theta=1-\sin^2\theta\), we have \[ -1+\cos^2\theta=-1+(1-\sin^2\theta)=-\sin^2\theta. \] 2. **Simplify the denominator:** - For \(\cos(-\theta)\), note that cosine is even: \[ \cos(-\theta)=\cos\theta. \] - For \(\sin(90^\circ+\theta)\), use the identity: \[ \sin(90^\circ+\theta)=\cos\theta. \] - For \(\tan^2\left(540^\circ+\theta\right)\), note that the tangent function has period \(180^\circ\). Since \[ 540^\circ=3\cdot180^\circ, \] we have \[ \tan(540^\circ+\theta)=\tan\theta, \] so \[ \tan^2(540^\circ+\theta)=\tan^2\theta. \] Combining these, the denominator becomes \[ \cos\theta\cdot\cos\theta\cdot\tan^2\theta=\cos^2\theta\,\tan^2\theta. \] Express \(\tan^2\theta\) in terms of sine and cosine: \[ \tan^2\theta=\frac{\sin^2\theta}{\cos^2\theta}. \] Hence, the denominator simplifies to \[ \cos^2\theta\cdot\frac{\sin^2\theta}{\cos^2\theta}=\sin^2\theta. \] 3. **Combine numerator and denominator:** The entire expression is then \[ \frac{-\sin^2\theta}{\sin^2\theta}=-1. \] --- **Final Answers:** - \(6.1.1:\quad \sin x=-\dfrac{3}{5}\) - \(6.1.2:\quad 2-\sin2x=\dfrac{26}{25}\) - \(6.1.3:\quad \cos^2(90^\circ-x)-1=-\dfrac{16}{25}\) - \(6.2:\quad -1\)