HSC BOARD – IMPORTANT FORMULAS
MATHS IMPORTANT FORMULAS FOR ALL STUDENTS.
1) Here is a list of Algebraic formulas –
- a2 – b2 = (a – b)(a + b)
- (a+b)2 = a2 + 2ab + b2
- a2 + b2 = (a – b)2 + 2ab
- (a – b)2 = a2 – 2ab + b2
- (a + b + c)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc
- (a – b – c)2 = a2 + b2 + c2 – 2ab – 2ac + 2bc
- (a + b)3 = a3 + 3a2b + 3ab2 + b3 ; (a + b)3 = a3 + b3 + 3ab(a + b)
- (a – b)3 = a3 – 3a2b + 3ab2 – b3
- a3 – b3 = (a – b)(a2 + ab + b2)
- a3 + b3 = (a + b)(a2 – ab + b2)
- (a + b)3 = a3 + 3a2b + 3ab2 + b3
- (a – b)3 = a3 – 3a2b + 3ab2 – b3
- (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4)
- (a – b)4 = a4 – 4a3b + 6a2b2 – 4ab3 + b4)
- a4 – b4 = (a – b)(a + b)(a2 + b2)
- a5 – b5 = (a – b)(a4 + a3b + a2b2 + ab3 + b4)
- If n is a natural number an – bn = (a – b)(an-1 + an-2b+…+ bn-2a + bn-1)
- If n is even (n = 2k), an + bn = (a + b)(an-1 – an-2b +…+ bn-2a – bn-1)
- If n is odd (n = 2k + 1), an + bn = (a + b)(an-1 – an-2b +…- bn-2a + bn-1)
- (a + b + c + …)2 = a2 + b2 + c2 + … + 2(ab + ac + bc + ….)
- Laws of Exponents (am)(an) = am+n (ab)m = ambm (am)n = amn
- Fractional Exponents a0 = 1 aman=am−n am = 1a−m a−m = 1am
- Roots of Quadratic Equation
- For a quadratic equation ax2 + bx + c where a ≠ 0, the roots will be given by the equation as −b±b2−4ac√2a
- Δ = b2 − 4ac is called the discrimination
- For real and distinct roots, Δ > 0
- For real and coincident roots, Δ = 0
- For non-real roots, Δ < 0
- If α and β are the two roots of the equation ax2 + bx + c then, α + β = (-b / a) and α × β = (c / a).
- If the roots of a quadratic equation are α and β, the equation will be (x − α)(x − β) = 0
- n! = (1).(2).(3)…..(n − 1).n
- n! = n(n − 1)! = n(n − 1)(n − 2)! = ….
- 0! = 1
- (a+b)n=an+nan−1b+n(n−1)2!an−2b2+n(n−1)(n−2)3!an−3b3+….+bn,where,n>1
Solved Examples
Question 1: Find out the value of 52 – 32Solution:
Using the formula a2 – b2 = (a – b)(a + b)
where a = 5 and b = 3
(a – b)(a + b)= (5 – 3)(5 + 3)= 2 × 8= 16
Question 2: 43 × 42 = ?
Solution:Using the exponential formula (am)(an) = am+nwhere a = 4
43 × 42= 43+2= 45= 1024
2) Here is a list of important Trigonometric formulas –
Quotient & Reciprocal Identities
tanθcscθ=sinθcosθcotθ=cosθsinθ=1sinθ secθ=1cosθ cotθ=1tanθ
Pythagorean Identities
sin2θ+cos2θ=11+cot2θ=csc2θtan2θ+1=sec2θ
Even & Odd Identities
sin(−x)csc(−x)=−sinx=−cscxcos(−x)=cosxsec(−x)=secxtan(−x)=−tanxcot(−x)=−cotx
Co-Function Identities
sin(π2−θ)=cosθcsc(π2−θ)=secθcos(π2−θ)=sinθsec(π2−θ)=cscθtan(π2−θ)=cotθcot(π2−θ)=tanθ
Sum and Difference Identities
cos(α+β)sin(α+β)tan(α+β)=cosαcosβ−sinαsinβ=sinαcosβ+cosαsinβ=tanα+tanβ1−tanαtanβcos(α−β)=cosαcosβ+sinαsinβsin(α−β)=sinαcosβ−cosαsinβtan(α−β)=tanα−tanβ1+tanαtanβ
Double Angle Identities
cos(2α)sin(2α)tan(2α)=cos2α−sin2α=2cos2α−1=1−2sin2α=2sinαcosβ=2tanα1−tan2α
Half Angle Identities
cosα2=±1+cosα2−−−−−−−−√sinα2=±1−cosα2−−−−−−−−√tanα2=1−cosαsinα=sinα1+cosα
Product to Sum & Sum to Product Identities
sina+sinbsina−sinbcosa+cosbcosa−cosb=2sina+b2cosa−b2=2sina−b2cosa+b2=2cosa+b2cosa−b2=2−2sina+b2sina−b2sinasinb=12[cos(a−b)−cos(a+b)]cosacosb=12[cos(a−b)+cos(a+b)]sinacosb=12[sin(a+b)+sin(a−b)]cosasinb=12[sin(a+b)−sin(a−b)]
Linear Combination Formula
Acosx+Bsinx=Ccos(x−D), where C=A2+B2−−−−−−−√,cosD=AC and sinD=BC
Review Questions
- Find the sine, cosine, and tangent of an angle with terminal side on (−8,15).
- If sina=5–√3 and tana<0, find seca.
- Simplify: cos4x−sin4xcos2x−sin2x.
- Verify the identity: 1+sinxcosxsinx=secx(cscx+1)
For problems 5-8, find all the solutions in the interval [0,2π).
- sec(x+π2)+2=0
- 8sin(x2)−8=0
- 2sin2x+sin2x=0
- 3tan22x=1
- Solve the trigonometric equation 1−sinx=3–√sinx over the interval [0,π].
- Solve the trigonometric equation 2cos3x−1=0 over the interval [0,2π].
- Solve the trigonometric equation 2sec2x−tan4x=3 for all real values of x.
Find the exact value of:
- cos157.5∘
- sin13π12
- Write as a product: 4(cos5x+cos9x)
- Simplify: cos(x−y)cosy−sin(x−y)siny
- Simplify: sin(4π3−x)+cos(x+5π6)
- Derive a formula for sin6x.
- If you solve cos2x=2cos2x−1 for cos2x, you would get cos2x=12(cos2x+1). This new formula is used to reduce powers of cosine by substituting in the right part of the equation for cos2x. Try writing cos4x in terms of the first power of cosine.
- If you solve cos2x=1−2sin2x for sin2x, you would get sin2x=12(1−cos2x). Similar to the new formula above, this one is used to reduce powers of sine. Try writing sin4x in terms of the first power of cosine.
- Rewrite in terms of the first power of cosine:
- sin2xcos22x
- tan42x
