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| source: ncert |
| topic: trigonometry |
| class_level: class_11 |
| chapter: ch3 |
| difficulty: intermediate |
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| Class 11 | Ch3: Trigonometric Functions |
| RADIAN MEASURE: pi radians=180°. 1 radian=180°/pi≈57.3°. |
| Arc length l=r×theta (theta in radians). Area of sector=(1/2)r^2×theta. |
| SIGN OF TRIG FUNCTIONS BY QUADRANT (ASTC rule: All Students Take Calculus): |
| Q1: all positive. Q2: sin,cosec positive. Q3: tan,cot positive. Q4: cos,sec positive. |
| TRIG VALUES FOR STANDARD ANGLES: (same as Class 10 table). |
| TRIG FUNCTIONS OF ALLIED ANGLES: |
| sin(pi-x)=sin(x). cos(pi-x)=-cos(x). tan(pi-x)=-tan(x). |
| sin(pi+x)=-sin(x). cos(pi+x)=-cos(x). tan(pi+x)=tan(x). |
| sin(2pi-x)=-sin(x). cos(2pi-x)=cos(x). |
| sin(-x)=-sin(x) (odd). cos(-x)=cos(x) (even). |
| SUM AND DIFFERENCE FORMULAS: |
| sin(A+B)=sinA cosB+cosA sinB. sin(A-B)=sinA cosB-cosA sinB. |
| cos(A+B)=cosA cosB-sinA sinB. cos(A-B)=cosA cosB+sinA sinB. |
| tan(A+B)=(tanA+tanB)/(1-tanA tanB). tan(A-B)=(tanA-tanB)/(1+tanA tanB). |
| DOUBLE ANGLE FORMULAS: |
| sin2A=2sinA cosA. cos2A=cos^2A-sin^2A=1-2sin^2A=2cos^2A-1. |
| tan2A=2tanA/(1-tan^2A). |
| HALF ANGLE: sin^2(A/2)=(1-cosA)/2. cos^2(A/2)=(1+cosA)/2. |
| PRODUCT TO SUM: |
| sinA cosB=(1/2)[sin(A+B)+sin(A-B)]. |
| cosA cosB=(1/2)[cos(A+B)+cos(A-B)]. |
| GENERAL SOLUTION: |
| sinx=sina → x=n*pi+(-1)^n*a. cosx=cosa → x=2n*pi±a. tanx=tana → x=n*pi+a. |
| SOLVED EXAMPLES: |
| Example 1: Prove sin(A+B)sin(A-B)=sin^2A-sin^2B. |
| LHS=(sinA cosB+cosA sinB)(sinA cosB-cosA sinB)=sin^2A cos^2B-cos^2A sin^2B. |
| =sin^2A(1-sin^2B)-(1-sin^2A)sin^2B=sin^2A-sin^2A sin^2B-sin^2B+sin^2A sin^2B=sin^2A-sin^2B. Proved. |
| Example 2: Solve 2cos^2x-3cosx+1=0 for x in [0,2pi]. |
| Let c=cosx: 2c^2-3c+1=0 → (2c-1)(c-1)=0 → c=1/2 or c=1. |
| cosx=1/2 → x=pi/3 or 5pi/3. cosx=1 → x=0. Solutions: {0, pi/3, 5pi/3}. |
| COMMON MISTAKES: |
| sinx=sin(a) general solution: x=n*pi+(-1)^n*a (alternating sign, NOT just ±a). |
| sin(A+B)≠sinA+sinB. Must use the formula. |
