Trigonometry deals with angles, triangles, and their ratios. It’s a fundamental part of JEE.
\(\sin \theta = \frac{b}{h}\) \(\cos \theta = \frac{b}{h}\) \(\tan \theta = \frac{b}{h}\)
\(\sin(90 + \theta) = -\cos \theta\) \(\cos(90 - \theta) = \sin \theta\)
\(\sin 30^\circ = \frac{1}{2}\) \(\sin 45^\circ = \frac{1}{\sqrt{2}}\) \(\sin 60^\circ = \frac{\sqrt{3}}{2}\) \(\sin 90^\circ = 1\)
| Col 1 | Col 2 | Col 3 |
|---|---|---|
sin²θ + cos²θ = 1 |
1 + cot²θ = cosec²θ |
1 + tan²θ = sec²θ |
sin (90 + θ) = cos θ |
sin (180 - θ) = sin θ |
sin (90 - θ) = cos θ |
cos (90 + θ) = - sin θ |
cos (180 - θ) = - cos θ |
cos (90 - θ) = sin θ |
Q: If sinθ = 3/5 and θ is in the first quadrant, find cosθ and tanθ.
Solution:
sinθ = 3/5 ⇒ Opposite = 3, Hypotenuse = 5cosθ = 4/5, tanθ = 3/4| Col 1 | Col 2 | Col 3 |
|---|---|---|
sin(360+θ) = sinθ |
cos(360+θ) = cosθ |
cos(360-θ) = cosθ |
sin(A+B) = sinA·cosB + sinB·cosA |
cos(A+B) = cosA·cosB - sinA·sinB |
sin(2A) = 2sinA·cosA |
1+cos(2A) = 2cos²(A) |
1-cos(2A) = 2sin²(A) |
tan (90 + θ) = - cot θ |
Straight lines are part of coordinate geometry and deal with equations of lines in a plane.
y = mx + cy - y₁ = m(x - x₁)Ax + By + C = 0Q: Find the equation of the line passing through (2, 3) with slope 4.
Solution:
Use point-slope form:
y - 3 = 4(x - 2)
⇒ y = 4x - 5
Differentiation is the rate of change of a function. It’s essential for calculus-based problems in JEE. Say (displacement) and $x_1, x_2$ time
$V_{av} = \langle v \rangle = \frac{y_2 - y_1}{t_2 - t_1} = 0$.
as the time interval $(t_2 - t_1)$ becomes very small the average velocity becomes instantaneous velocity.
so $V_{av}$ changes to $v$; $\Delta v / \Delta t$ becomes closer to $x_1, x_2$ line.
which was intersecting the curve at two points would eventually touch at just at a single point
d/dx(xⁿ) = nxⁿ⁻¹d/dx(sin x) = cos xd/dx(eˣ) = eˣd/dx(ln x) = 1/xQ: Differentiate f(x) = x³ + 2x² - 5x + 1
Solution:
f'(x) = 3x² + 4x - 5
Integration is the reverse of differentiation. It’s used to calculate area, displacement, and more.
∫xⁿ dx = xⁿ⁺¹ / (n+1) + C, n ≠ -1∫1/x dx = ln|x| + C∫eˣ dx = eˣ + C∫cos x dx = sin x + CQ: Find ∫(3x² + 2x - 1) dx
Solution:
∫(3x² + 2x - 1) dx = x³ + x² - x + C
Equal Vectors $\vec{A} = \vec{B}$ If two vectors are equal then their magnitudes and directions are identical,
Multiply vector by a scalar
$\vec{B} = 2\vec{A}$
$\vec{A}$ + $\vec{A}$ = $\vec{B}$
When two vectors are arranged such that the need of one vector coincides with the tail of the other, then the vector obtained in completing the triangle, in the direction from tail to head is the sum of the two vectors.
Practice these concepts regularly with mock papers and past year questions. Tricky problems often require blending concepts across these topics.