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| source: ncert |
| topic: trigonometry |
| class_level: jee_advanced |
| chapter: advanced_trigonometry |
| difficulty: advanced |
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| JEE Advanced | Topic 4: Advanced Trigonometry |
| PRODUCT FORMULAS: |
| sin A + sin B = 2 sin((A+B)/2) cos((A-B)/2). |
| sin A - sin B = 2 cos((A+B)/2) sin((A-B)/2). |
| cos A + cos B = 2 cos((A+B)/2) cos((A-B)/2). |
| cos A - cos B = -2 sin((A+B)/2) sin((A-B)/2). |
| sin A * sin B = (1/2)[cos(A-B)-cos(A+B)]. |
| cos A * cos B = (1/2)[cos(A-B)+cos(A+B)]. |
| sin A * cos B = (1/2)[sin(A+B)+sin(A-B)]. |
| TRIGONOMETRIC EQUATIONS: |
| General solutions: sin(theta)=sin(alpha) → theta=n*pi+(-1)^n*alpha. |
| cos(theta)=cos(alpha) → theta=2n*pi±alpha. |
| tan(theta)=tan(alpha) → theta=n*pi+alpha. |
| For sin^2(theta)=sin^2(alpha): theta=n*pi±alpha. |
| CONDITIONAL IDENTITIES (when A+B+C=pi, i.e., angles of triangle): |
| sin2A+sin2B+sin2C=4sinA sinB sinC. |
| sinA+sinB+sinC=4cos(A/2)cos(B/2)cos(C/2). |
| cos A+cosB+cosC=1+4sin(A/2)sin(B/2)sin(C/2). |
| tan A+tanB+tanC=tanA tanB tanC (since A+B+C=pi). |
| INVERSE TRIG ADVANCED: |
| sin^(-1)x+sin^(-1)y=sin^(-1)(x*sqrt(1-y^2)+y*sqrt(1-x^2)) if x^2+y^2<=1. |
| tan^(-1)x+tan^(-1)y=pi+tan^(-1)((x+y)/(1-xy)) if x>0, y>0, xy>1. |
| TRIGONOMETRIC INEQUALITIES: |
| sin x > sin a for x in (a, pi-a) when a in (0, pi/2). |
| Solve by graphical method: draw y=sinx and horizontal line y=sin(a). |
| SOLVED EXAMPLES: |
| Example 1: Prove cos(pi/7)*cos(2pi/7)*cos(3pi/7)=1/8. |
| Use sin(2^n*A)=2^n*sinA*cosA*cos(2A)*... |
| sin(8pi/7)=sin(pi+pi/7)=-sin(pi/7). |
| 2^3*sin(pi/7)*cos(pi/7)*cos(2pi/7)*cos(4pi/7)=sin(8pi/7)=-sin(pi/7). |
| 8*sin(pi/7)*cos(pi/7)*cos(2pi/7)*cos(4pi/7)=-sin(pi/7). |
| cos(pi/7)*cos(2pi/7)*cos(4pi/7)=-1/8. Note cos(4pi/7)=cos(pi-3pi/7)=-cos(3pi/7). |
| So cos(pi/7)*cos(2pi/7)*(-(-cos(3pi/7)))... gives final answer 1/8. |
| Example 2: Solve sin2x-sinx=cosx-cos2x for x in [0,2pi]. |
| sin2x+cos2x=sinx+cosx. 2sinx cosx+2cos^2x-1=sinx+cosx. |
| Let s=sinx, c=cosx. 2sc+2c^2-s-c-1=0. Factor: (2c-1)(c+s)... |
| Try: (sinx+cosx)(2cosx-1)-(sinx+cosx)=0... |
| (sinx+cosx-1)(2cosx-1)=0. |
| Case 1: sinx+cosx=1 → sqrt(2)sin(x+pi/4)=1 → x+pi/4=pi/4 or 3pi/4 → x=0 or pi/2. |
| Case 2: cosx=1/2 → x=pi/3 or 5pi/3. |
| Solutions: {0, pi/3, pi/2, 5pi/3}. |
| COMMON MISTAKES: |
| Sum-to-product and product-to-sum: memorise both directions. |
| Conditional identities: only valid when A+B+C=pi (triangle angles). |
| General solution: write COMPLETE general solution then find specific values in given interval. |
