Saved Bookmarks
| 1. |
Evaluate cos105degree+sin105degree |
| Answer» The value of cos 105o\xa0+ sin 105o\xa0is\xa0{tex}\\frac{1}{\\sqrt{2}}{/tex}Cos trigonometric values are positive in “first and fourth” quadrants in coordinate axes.Sin trigonometric values are positive in “first and second” quadrants in coordinate axes.cos 105o\xa0+ sin 105o\xa0= cos (90o\xa0+ 15o) + sin (90o\xa0+ 15o)= - sin (15o) + cos (15o) (ascos (90o\xa0+\xa0{tex}\\theta{/tex}) = - sin\xa0{tex}\\theta{/tex}\xa0and sin (90o +\xa0{tex}\\theta{/tex}) = cos\xa0{tex}\\theta{/tex})= cos 15o - sin 15o{tex}\\left.=\\sqrt{2}\\left[\\frac{1}{\\sqrt{2}} \\cos 15^{\\circ}-\\frac{1}{\\sqrt{2}} \\sin 15^{\\circ}\\right] \\quad \\text { (Multiply and divide by } \\sqrt{2}\\right){/tex}{tex}=\\sqrt{2}\\left[\\sin 45^{\\circ} \\cos 15^{\\circ}-\\cos 45^{\\circ} \\sin 15^{\\circ}\\right] \\quad\\left(A s \\sin 45^{\\circ}=\\cos 45^{\\circ}=\\frac{1}{\\sqrt{2}}\\right){/tex}{tex}=\\sqrt{2}\\left[\\sin \\left(45^{\\circ}-15^{\\circ}\\right)\\right] \\quad(A \\sin (A-B)=\\sin A \\cos B-\\cos A \\sin B){/tex}{tex}=\\sqrt{2}\\left[\\sin 30^{\\circ}\\right]{/tex}{tex}=\\sqrt{2} \\times \\frac{1}{2}=\\frac{1}{\\sqrt{2}}{/tex} | |